{"id":3912,"date":"2022-04-05T18:32:27","date_gmt":"2022-04-05T18:32:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3912"},"modified":"2022-10-29T01:13:48","modified_gmt":"2022-10-29T01:13:48","slug":"limits","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/limits\/","title":{"raw":"Limits","rendered":"Limits"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the limit of a function of two variables.<\/li>\r\n \t<li>Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach.<\/li>\r\n<\/ul>\r\n<\/div>\r\nRecall from <a href=\"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/definition-of-a-limit\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"8b89d172-2927-466f-8661-01abc7ccdba4\" data-page-slug=\"2-2-the-limit-of-a-function\">The Limit of a Function<\/a> the definition of a limit of a function of one variable:\r\n\r\nLet [latex]f\\,(x)[\/latex]\u00a0be defined for all [latex]x\\neq{a}[\/latex]\u00a0in an open interval containing [latex]a[\/latex]. Let [latex]L[\/latex] be a real number. Then\r\n<p style=\"text-align: center;\">[latex]\\displaystyle{\\lim_{x\\to{a}}}f\\,(x)=L[\/latex]<\/p>\r\nif for every [latex]\\epsilon\\,&gt;\\,0[\/latex], there exists a [latex]\\delta\\,&gt;\\,0[\/latex], such that if [latex]0\\,&lt;\\,|x-a|\\,&lt;\\delta[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex], then\r\n<p style=\"text-align: center;\">[latex]|f\\,(x)-L|\\,&lt;\\,\\epsilon[\/latex]<\/p>\r\n<p id=\"fs-id1167794031084\" class=\" \">Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend the idea of an open interval in one variable to an open interval in two variables.<\/p>\r\n\r\n<div id=\"fs-id1167793269223\" class=\"ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nConsider a point [latex](a,\\ b)\\in\\mathbb{R}^2[\/latex]. A [latex]\\delta[\/latex] <strong>disk<\/strong>\u00a0centered at point [latex](a,\\ b)[\/latex] is defined\u00a0to be an open disk of radius [latex]\\delta[\/latex] centered at point\u00a0[latex](a,\\ b)[\/latex] - that is,\r\n<p style=\"text-align: center;\">[latex]\\{(x,\\ y)\\in\\mathbb{R}^{2}|(x-a)^{2}+(y-b)^{2}&lt;\\,\\delta^{2}\\}[\/latex]<\/p>\r\nas shown in the following graph.\r\n\r\n<\/div>\r\n[caption id=\"attachment_970\" align=\"aligncenter\" width=\"423\"]<img class=\"size-full wp-image-970\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28223402\/4-2-1.jpeg\" alt=\"On the xy plane, the point (2, 1) is shown, which is the center of a circle of radius \u03b4.\" width=\"423\" height=\"197\" \/> Figure 1. A [latex]\\delta[\/latex] disk centered around the point [latex](2,1)[\/latex].[\/caption]The idea of a [latex]\\delta[\/latex] disk\u00a0appears in the definition of the limit of a function of two variables. If [latex]\\delta[\/latex]\u00a0is small, then all the points [latex](x,\\ y)[\/latex] in the [latex]\\delta[\/latex] disk are close to [latex](a,\\ b)[\/latex].\u00a0This is completely analogous to [latex]x[\/latex] being close to [latex]a[\/latex]\u00a0in the definition of a limit of a function of one variable. In one dimension, we express this restriction as\r\n<p style=\"text-align: center;\">[latex]a-\\delta\\,&lt;\\,x\\,&lt;\\,a+\\delta[\/latex].<\/p>\r\nIn more than one dimension, we use a [latex]\\delta[\/latex] disk.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]f[\/latex]\u00a0be a function of two variables, [latex]x[\/latex] and [latex]y[\/latex]. The limit of [latex]f\\,(x,\\ y)[\/latex] as [latex](x,\\ y)[\/latex] approaches [latex](a,\\ b)[\/latex] is [latex]L[\/latex], written\r\n<div style=\"text-align: center;\">[latex]\\large{\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}f\\,(x,\\ y)=L}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nif for each [latex]\\epsilon&gt;0[\/latex]\u00a0there exists a small enough [latex]\\delta\\,&gt;\\,0[\/latex]\u00a0such that for all points [latex](x,\\ y)[\/latex] in a [latex]\\delta[\/latex] disk around [latex](a,\\ b)[\/latex], except possibly for [latex](a,\\ b)[\/latex] itself, the value of [latex]f\\,(x,\\ y)[\/latex] is no more than [latex]\\epsilon[\/latex] away from [latex]L[\/latex]\u00a0(Figure 2). Using symbols, we write the following: For any [latex]\\varepsilon\\,&gt;\\,0[\/latex], there exists a number [latex]\\delta\\,&gt;\\,0[\/latex] such that\r\n<div style=\"text-align: center;\">[latex]\\large{|f\\,(x,\\ y)-L|\\,&lt;\\,\\varepsilon \\text{ whenever }0\\,&lt;\\,\\sqrt{(x-a)^{2}+(y-b)^{2}}\\,&lt;\\,\\delta}[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[caption id=\"attachment_971\" align=\"aligncenter\" width=\"646\"]<img class=\"size-full wp-image-971\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28223617\/4-2-2.jpeg\" alt=\"In xyz space, a function is drawn with point L. This point L is the center of a circle of radius \u0949, with points L \u00b1 \u0949 marked. On the xy plane, there is a point (a, b) drawn with a circle of radius \u03b4 around it. This is denoted the \u03b4-disk. There are dashed lines up from the \u03b4-disk to make a disk on the function, which is called the image of delta disk. Then there are dashed lines from this disk to the circle around the point L, which is called the \u0949-neighborhood of L.\" width=\"646\" height=\"581\" \/> Figure 2.\u00a0The limit of a function involving two variables requires that\u00a0[latex]f(x,y)[\/latex] be within\u00a0[latex]\\varepsilon[\/latex] of\u00a0[latex]L[\/latex]\u00a0whenever [latex](x,y)[\/latex]\u00a0is within [latex]\\delta[\/latex] of\u00a0[latex](a,b)[\/latex]. The smaller the value of\u00a0[latex]\\varepsilon[\/latex],\u00a0the smaller the value of\u00a0[latex]\\delta[\/latex].[\/caption]Proving that a limit exists using the definition of a <span id=\"term158\" class=\"no-emphasis\" data-type=\"term\">limit of a function of two variables<\/span> can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of the formulas in the limit laws theorem in <a href=\"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/introduction-to-the-limit-laws\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"8b89d172-2927-466f-8661-01abc7ccdba4\" data-page-slug=\"2-3-the-limit-laws\">The Limit Laws<\/a>.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Limit LAws for Functions of Two Variables Theorem<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]f\\,(x,\\ y)[\/latex]\u00a0and [latex]g\\,(x,\\ y)[\/latex] be defined for all [latex](x,\\ y)\\neq(a,\\ b)[\/latex] in a neighborhood around [latex](a,\\ b)[\/latex], and assume\u00a0the neighborhood is contained completely inside the domain of [latex]f[\/latex]. Assume that [latex]L[\/latex] and [latex]M[\/latex] are real numbers such that [latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}f\\,(x,\\ y)=L[\/latex] and [latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}g\\,(x,\\ y)=M[\/latex], and let [latex]c[\/latex] be a constant.\u00a0Then each of the following statements holds:\r\n\r\n<strong>Constant\u00a0Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}c=c[\/latex]<\/p>\r\n<strong>Identity Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}x} &amp; =\\hfill &amp; {a} \\hfill \\\\ \\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}y} &amp; =\\hfill &amp; {b} \\end{array}[\/latex]<\/p>\r\n<strong>Sum Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)+g\\,(x,\\ y))=L+M[\/latex]<\/p>\r\n<strong>Difference Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)-(g\\,(x,\\ y))=L-M[\/latex]<\/p>\r\n<strong>Constant Multiple<\/strong> <strong>Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(cf\\,(x,\\ y))=cL[\/latex]<\/p>\r\n<strong>Product Law:\u00a0<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)\\,g\\,(x,\\ y))=LM[\/latex]<\/p>\r\n<strong>Quotient Law:\u00a0<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\frac{f\\,(x,\\ y)}{g\\,(x,\\ y)}=\\frac{L}{M}[\/latex] for [latex]M\\neq{0}[\/latex]<\/p>\r\n<strong>Power Law:\u00a0<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y))^{n}=L^{n}[\/latex]<\/p>\r\nfor any positive integer [latex]n[\/latex].\r\n\r\n<strong>Root Law:<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\sqrt[n]{f\\,(x,\\ y)}=\\sqrt[n]{L}[\/latex]<\/p>\r\nfor all [latex]L[\/latex] if [latex]n[\/latex] is odd and positive, and for [latex]L\\,\\geq\\,0[\/latex] if [latex]n[\/latex] is even and positive.\r\n\r\n<\/div>\r\nThe proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Limit of a Function of Two Variables<\/h3>\r\nFind each of the following limits:\r\n<ol>\r\n \t<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}\\frac{2x+3y}{4x-3y}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1167793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793933114\"]\r\n<ol>\r\n \t<li>First use the sum and difference laws to separate the terms:<\/li>\r\n<\/ol>\r\n[latex]\\hspace{5cm}\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\hspace{5cm}=\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x^{2}\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}2xy\\bigg) +\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y^{2}\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}4x\\bigg) \\\\ \\hspace{5cm}+\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNext,\u00a0use the constant multiple law on the second, third, fourth, and fifth limits:\r\n<div style=\"text-align: center;\">[latex]\\hspace{5cm}=\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x^{2}\\bigg)-2\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}xy\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y^{2}\\bigg)-4\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg) \\\\ \\hspace{5cm} +3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNow, use the power law on the first and third limits, and the product law on the second limit:\r\n<div style=\"text-align: center;\">[latex]\\hspace{5cm} =\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg)^{2}-2\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}xy\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)^{2} \\\\ \\hspace{5cm} -4\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\r\n&nbsp;\r\n\r\nLast, use the identity laws on the first six limits and the constant law on the last limit:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)}&amp; =\\hfill &amp; {(2)^{2}-2(2)(-1)+3(-1)^{2}-4(2)+3(-1)-6} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {-6} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n<ol>\r\n \t<li>Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(4x-3y)} &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}4x-\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {4\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\big)-3\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\big)} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {4(2)-3(-1)=11.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nSince the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(2x+3y)} &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}2x+\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\big)+3\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\big)} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {2(2)+3(-1)}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {1.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nTherefore, according to the quotient law we have\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}\\frac{2x+3y}{4x-3y}=\\frac{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(2x+3y)}{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(4x-3y)}=\\frac{1}{11}.[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nEvaluate the following limit:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(5,\\ -2)}\\sqrt[3]{\\frac{x^{2}-y}{y^{2}+x-1}.}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1367793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1367793933114\"]\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(5,\\ -2)}\\sqrt[3]{\\frac{x^{2}-y}{y^{2}+x-1}}=\\frac{3}{2}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSince we are taking the limit of a function of two variables, the point [latex](a,\\ b)[\/latex] is in [latex]\\mathbb{R}^{2}[\/latex],\u00a0and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward\u00a0[latex](a,\\ b)[\/latex].\u00a0If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Limits that Fail to Exist<\/h3>\r\nShow that neither of the following limits exist:\r\n<ol>\r\n \t<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{2xy}{3x^{2}+y^{2}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1197793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1197793933114\"]\r\n<ol>\r\n \t<li>The domain of the function [latex]f\\,(x,\\ y)=\\frac{2xy}{3x^{2}+y^{2}}[\/latex]\u00a0consists of all points in the [latex]xy[\/latex]-plane\u00a0except for the point [latex](0,\\ 0)[\/latex]\u00a0(Figure 3). To show that the limit does not exist as [latex](x,\\ y)[\/latex] approaches [latex](0,\\ 0)[\/latex],\u00a0we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point [latex](0,\\ 0)[\/latex].\u00a0First, consider the line [latex]y=0[\/latex] in the [latex]xy[\/latex]-plane. Substituting\u00a0[latex]y=0[\/latex] into [latex]f\\,(x,\\ y)[\/latex] gives\r\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ 0)=\\frac{2x(0)}{3x^{2}+0^{2}}=0}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nFor any value of [latex]x[\/latex].\u00a0Therefore the value of [latex]f[\/latex]\u00a0remains constant for any point on the [latex]x[\/latex]-axis, and as [latex]y[\/latex] approaches zero, the function remains fixed at zero.\u00a0Next, consider the line [latex]y=x[\/latex]. Substituting [latex]y=x[\/latex] into [latex]f\\,(x,\\ y)[\/latex] gives\r\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ x)=\\frac{2x(x)}{3x^{2}+x^{2}}=\\frac{2x^{2}}{4x^{2}}=\\frac{1}{2}}.[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThis is true for any point on the line [latex]y=x[\/latex]. If we let [latex]x[\/latex] approach zero\u00a0while staying on this line, the value of the function remains fixed at [latex]\\frac{1}{2}[\/latex],\u00a0regardless of how small [latex]x[\/latex] is.\u00a0Choose a value for [latex]\\epsilon[\/latex] that is less than [latex]1\/2-[\/latex]say, [latex]1\/4[\/latex]. Then, no matter how small a [latex]\\delta[\/latex] disk\u00a0we draw around [latex](0,\\ 0)[\/latex], the values of [latex]f\\,(x,\\ y)[\/latex] for points inside that [latex]\\delta[\/latex] disk will include both [latex]0[\/latex] and [latex]\\frac{1}{2}.[\/latex] Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.\r\n\r\n[caption id=\"attachment_973\" align=\"aligncenter\" width=\"451\"]<img class=\"size-full wp-image-973\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28224845\/4-2-3.jpeg\" alt=\"In xyz space, the function f(x, y) = 2xy\/(3x2 + y2) is shown, which is a slightly twisted plane, with values of 0 along the line y = 0 and values of \u00bd along the line y = x.\" width=\"451\" height=\"288\" \/> Figure 3. Graph of the function [latex]\\small{f(x,y)=(2xy)\/(3x^{2}+y^{2})}[\/latex]. Along the line [latex]\\small{y=0}[\/latex], the function is equal to zero; along the line [latex]\\small{y=x}[\/latex], the function is equal to [latex]\\small{\\frac{1}{2}}[\/latex].[\/caption]<\/li>\r\n \t<li>In a similar fashion to 1., we can approach the origin along any straight line passing through the origin. If we try the [latex]x[\/latex]-axis (i.e., [latex]y=0[\/latex]),\u00a0then the function remains fixed at zero. The same is true for the [latex]y[\/latex]-axis.\u00a0Suppose we approach the origin along a straight line of slope [latex]k[\/latex]\u00a0The equation of this line is [latex]y=kx[\/latex]. Then the limit becomes\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4x(kx)^{2}}{x^{2}+3(kx)^{4}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4k^{2}x^{3}}{x^{2}+3k^{4}x^{4}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4k^{2}x}{1+3k^{4}x^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\frac{\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}4k^{2}x}{\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}(1+3k^{4}x^{2})}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {0}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nRegardless of the value of [latex]k[\/latex].\u00a0It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation [latex]x=y^{2}[\/latex]. Substituting [latex]y^{2}[\/latex] in place of [latex]x[\/latex] in [latex]f\\,(x,\\ y)[\/latex] gives\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}} &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4(y^{2})y^{2}}{(y^{2})^{2}+3y^{4}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4y^{4}}{y^{4}+3y^{4}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}1}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {1.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nBy the same logic\u00a0in 1., it is impossible to find a [latex]\\delta[\/latex] disk\u00a0around the origin that satisfies the definition of the limit for any value of [latex]\\epsilon\\,&lt;\\,1[\/latex]. Therefore, [latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}[\/latex] does not exist.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nShow that\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ 1)}\\frac{(x-2)(y-1)}{(x-2)^{2}+(y-1)^{2}}[\/latex]<\/p>\r\ndoes not exist.\r\n\r\n[reveal-answer q=\"fs-id1667793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1667793933114\"]\r\n<div style=\"text-align: center;\">If [latex]y=k\\,(x-2)+1[\/latex], then [latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ 1)}\\frac{(x-2)(y-1)}{(x-2)^{2}+(y-1)^{2}}=\\frac{k}{1+k^{2}}[\/latex]. Since the answer depends on [latex]k[\/latex], the limit fails to exist.<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186150&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=pWvDsHUhTfY&amp;video_target=tpm-plugin-dfjtf1gg-pWvDsHUhTfY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.7_transcript.html\">transcript for \u201cCP 4.7\u201d here (opens in new window).<\/a><\/center>\r\n<h2>Interior Points and Boundary Points<\/h2>\r\nTo study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]S[\/latex] be a subset of [latex]\\mathbb{R}^{2}[\/latex]\u00a0(Figure 4).\r\n<ol>\r\n \t<li>A point [latex]P_0[\/latex] is called an\u00a0<strong>interior point<\/strong> of [latex]S[\/latex] if there is a [latex]\\delta[\/latex] disk centered around [latex]P_0[\/latex] contained completely in [latex]S[\/latex].<\/li>\r\n \t<li>A point [latex]P_0[\/latex] is called a\u00a0<strong>boundary point\u00a0<\/strong>of [latex]S[\/latex] if every [latex]\\delta[\/latex] disk centered around [latex]P_0[\/latex]\u00a0contains points both inside and outside [latex]S[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[caption id=\"attachment_974\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-974\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225116\/4-2-4.jpeg\" alt=\"On the xy plane, a closed shape is drawn. There is a point (\u20131, 1) drawn on the inside of the shape, and there is a point (2, 3) drawn on the boundary. Both of these points are the centers of small circles.\" width=\"417\" height=\"422\" \/> Figure 4.\u00a0In the set [latex]S[\/latex] shown,\u00a0[latex]\\small{(-1,1)}[\/latex] is an interior point and\u00a0[latex]\\small{(2,3)}[\/latex]\u00a0is a boundary point.[\/caption]\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]S[\/latex] be a subset of [latex]\\mathbb{R}^{2}[\/latex]\u00a0(Figure 4).\r\n<ol>\r\n \t<li>[latex]S[\/latex] is called an\u00a0<strong>open set<\/strong> if every point of [latex]S[\/latex] is an interior point.<\/li>\r\n \t<li>[latex]S[\/latex]\u00a0is called a <strong><span id=\"term162\" data-type=\"term\">closed set<\/span><\/strong> if it contains all its boundary points.<\/li>\r\n<\/ol>\r\n<\/div>\r\nAn example of an open set is a [latex]\\delta[\/latex]\u00a0disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a [latex]\\delta[\/latex]\u00a0disk but not the other half, then the set is neither open nor closed.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]S[\/latex] be a subset of [latex]\\mathbb{R}^{2}[\/latex]\u00a0(Figure 4).\r\n<ol>\r\n \t<li>An open set [latex]S[\/latex]\u00a0is a <strong><span id=\"term163\" data-type=\"term\">connected set<\/span> <\/strong>if it cannot be represented as the union of two or more disjoint, nonempty open subsets.<\/li>\r\n \t<li>A set [latex]S[\/latex]\u00a0is a <strong><span id=\"term164\" data-type=\"term\">region<\/span><\/strong> if it is open, connected, and nonempty.<\/li>\r\n<\/ol>\r\n<\/div>\r\nThe definition of a limit of a function of two variables requires the [latex]\\delta[\/latex]\u00a0disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the [latex]\\delta[\/latex] disk is not contained inside the domain. By definition, some of the points of the\u00a0[latex]\\delta[\/latex] disk\u00a0are inside the domain and some are outside. Therefore, we need only consider points that are inside both the [latex]\\delta[\/latex]\u00a0disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]f[\/latex] be a function of two variables, [latex]x[\/latex] and [latex]y[\/latex], and suppose [latex](a,\\ b)[\/latex]\u00a0is on the boundary of the domain of [latex]f[\/latex]. Then, the limit of [latex]f\\,(x,\\ y)[\/latex] as [latex](x,\\ y)[\/latex] approaches [latex](a,\\ b)[\/latex] is [latex]L[\/latex], written\r\n<div style=\"text-align: center;\">[latex]\\large{\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\,f\\,(x,\\ y)=L}[\/latex],<\/div>\r\n&nbsp;\r\n\r\nif for any [latex]\\epsilon\\,&gt;\\,0[\/latex], there exists a number [latex]\\delta\\,&gt;\\,0[\/latex] such that for any point [latex](x,\\ y)[\/latex] inside the domain of [latex]f[\/latex] and within a suitably small distance positive [latex]\\delta[\/latex] of [latex](a,\\ b)[\/latex], the value of [latex]f\\,(x,\\ y)[\/latex] is no more than [latex]\\epsilon[\/latex] away from [latex]L[\/latex]\u00a0(Figure 2). Using symbols, we can write: For any [latex]\\epsilon\\,&gt;\\,0[\/latex], there exists a number [latex]\\delta\\,&gt;\\,0[\/latex] such that\r\n<div style=\"text-align: center;\">[latex]\\large{|f\\,(x,\\ y)-L|\\,&lt;\\,\\epsilon}[\/latex] whenever [latex]\\large{0\\,&lt;\\,\\sqrt{(x-a)^{2}+(y-b)^{2}}\\,&lt;\\,\\delta}[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Limit of a Function at a Boundary Point<\/h3>\r\nProve [latex]\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}\\sqrt{25-x^{2}-y^{2}}=0[\/latex].\r\n\r\n[reveal-answer q=\"fs-id1667743933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1667743933114\"]\r\n<div style=\"text-align: left;\">The domain of the function [latex]f\\,(x,\\ y)=\\sqrt{25-x^{2}-y^{2}}[\/latex] is [latex]\\{(x,\\ y)\\in\\mathbb{R}^{2}\\,|\\,x^{2}+y^{2}\\,\\leq\\,25\\}[\/latex], which\u00a0is a circle of radius [latex]5[\/latex]\u00a0centered at the origin, along with its interior as shown in the following graph.<\/div>\r\n<div>[caption id=\"attachment_976\" align=\"aligncenter\" width=\"417\"]<img class=\"wp-image-976 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225404\/4-2-5.jpeg\" alt=\"A circle with radius 5 centered at the origin with its interior filled in.\" width=\"417\" height=\"422\" \/> Figure 5. Domain of the function [latex]\\small{f(x,y)=\\sqrt{25-x^{2}-y^{2}}}[\/latex].[\/caption]<\/div>\r\n<div>We can\u00a0use the limit laws, which apply to limits at the boundary of domains as well as interior points:<\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}\\sqrt{25-x^{2}-y^{2}}} &amp; =\\hfill &amp; {\\sqrt{\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}(25-x^{2}-y^{2})}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}25-\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}x^{2}-\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}y^{2}}} \\hfill \\\\ \\hfill &amp; =\\hfill &amp; {\\sqrt{25-4^{2}-3^{2}}}\\hfill \\\\ \\hfill &amp; =\\hfill &amp; {0.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\r\n<div>See the following graph.<\/div>\r\n<div>[caption id=\"attachment_977\" align=\"aligncenter\" width=\"353\"]<img class=\"size-full wp-image-977\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225533\/4-2-6.jpeg\" alt=\"The upper hemisphere in xyz space with radius 5 and center the origin.\" width=\"353\" height=\"378\" \/> Figure 6.\u00a0Graph of the function [latex]\\small{f(x,y)=\\sqrt{25-x^{2}-y^{2}}}[\/latex].[\/caption]<\/div>\r\n<div><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nEvaluate the following limit:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(5,-2)}\\sqrt{29-x^{2}-y^{2}}.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1667793733114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1667793733114\"]\r\n<div style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(5,-2)}\\sqrt{29-x^{2}-y^{2}}=0[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186151&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=zYJIT-NNVKM&amp;video_target=tpm-plugin-0y6dbuw0-zYJIT-NNVKM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.8_transcript.html\">transcript for \u201cCP 4.8\u201d here (opens in new window).<\/a><\/center>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the limit of a function of two variables.<\/li>\n<li>Learn how a function of two variables can approach different values at a boundary point, depending on the path of approach.<\/li>\n<\/ul>\n<\/div>\n<p>Recall from <a href=\"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/definition-of-a-limit\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"8b89d172-2927-466f-8661-01abc7ccdba4\" data-page-slug=\"2-2-the-limit-of-a-function\">The Limit of a Function<\/a> the definition of a limit of a function of one variable:<\/p>\n<p>Let [latex]f\\,(x)[\/latex]\u00a0be defined for all [latex]x\\neq{a}[\/latex]\u00a0in an open interval containing [latex]a[\/latex]. Let [latex]L[\/latex] be a real number. Then<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle{\\lim_{x\\to{a}}}f\\,(x)=L[\/latex]<\/p>\n<p>if for every [latex]\\epsilon\\,>\\,0[\/latex], there exists a [latex]\\delta\\,>\\,0[\/latex], such that if [latex]0\\,<\\,|x-a|\\,<\\delta[\/latex] for all [latex]x[\/latex] in the domain of [latex]f[\/latex], then\n\n\n<p style=\"text-align: center;\">[latex]|f\\,(x)-L|\\,<\\,\\epsilon[\/latex]<\/p>\n<p id=\"fs-id1167794031084\" class=\"\">Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend the idea of an open interval in one variable to an open interval in two variables.<\/p>\n<div id=\"fs-id1167793269223\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Consider a point [latex](a,\\ b)\\in\\mathbb{R}^2[\/latex]. A [latex]\\delta[\/latex] <strong>disk<\/strong>\u00a0centered at point [latex](a,\\ b)[\/latex] is defined\u00a0to be an open disk of radius [latex]\\delta[\/latex] centered at point\u00a0[latex](a,\\ b)[\/latex] &#8211; that is,<\/p>\n<p style=\"text-align: center;\">[latex]\\{(x,\\ y)\\in\\mathbb{R}^{2}|(x-a)^{2}+(y-b)^{2}<\\,\\delta^{2}\\}[\/latex]<\/p>\n<p>as shown in the following graph.<\/p>\n<\/div>\n<div id=\"attachment_970\" style=\"width: 433px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-970\" class=\"size-full wp-image-970\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28223402\/4-2-1.jpeg\" alt=\"On the xy plane, the point (2, 1) is shown, which is the center of a circle of radius \u03b4.\" width=\"423\" height=\"197\" \/><\/p>\n<p id=\"caption-attachment-970\" class=\"wp-caption-text\">Figure 1. A [latex]\\delta[\/latex] disk centered around the point [latex](2,1)[\/latex].<\/p>\n<\/div>\n<p>The idea of a [latex]\\delta[\/latex] disk\u00a0appears in the definition of the limit of a function of two variables. If [latex]\\delta[\/latex]\u00a0is small, then all the points [latex](x,\\ y)[\/latex] in the [latex]\\delta[\/latex] disk are close to [latex](a,\\ b)[\/latex].\u00a0This is completely analogous to [latex]x[\/latex] being close to [latex]a[\/latex]\u00a0in the definition of a limit of a function of one variable. In one dimension, we express this restriction as<\/p>\n<p style=\"text-align: center;\">[latex]a-\\delta\\,<\\,x\\,<\\,a+\\delta[\/latex].<\/p>\n<p>In more than one dimension, we use a [latex]\\delta[\/latex] disk.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]f[\/latex]\u00a0be a function of two variables, [latex]x[\/latex] and [latex]y[\/latex]. The limit of [latex]f\\,(x,\\ y)[\/latex] as [latex](x,\\ y)[\/latex] approaches [latex](a,\\ b)[\/latex] is [latex]L[\/latex], written<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}f\\,(x,\\ y)=L}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>if for each [latex]\\epsilon>0[\/latex]\u00a0there exists a small enough [latex]\\delta\\,>\\,0[\/latex]\u00a0such that for all points [latex](x,\\ y)[\/latex] in a [latex]\\delta[\/latex] disk around [latex](a,\\ b)[\/latex], except possibly for [latex](a,\\ b)[\/latex] itself, the value of [latex]f\\,(x,\\ y)[\/latex] is no more than [latex]\\epsilon[\/latex] away from [latex]L[\/latex]\u00a0(Figure 2). Using symbols, we write the following: For any [latex]\\varepsilon\\,>\\,0[\/latex], there exists a number [latex]\\delta\\,>\\,0[\/latex] such that<\/p>\n<div style=\"text-align: center;\">[latex]\\large{|f\\,(x,\\ y)-L|\\,<\\,\\varepsilon \\text{ whenever }0\\,<\\,\\sqrt{(x-a)^{2}+(y-b)^{2}}\\,<\\,\\delta}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"attachment_971\" style=\"width: 656px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-971\" class=\"size-full wp-image-971\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28223617\/4-2-2.jpeg\" alt=\"In xyz space, a function is drawn with point L. This point L is the center of a circle of radius \u0949, with points L \u00b1 \u0949 marked. On the xy plane, there is a point (a, b) drawn with a circle of radius \u03b4 around it. This is denoted the \u03b4-disk. There are dashed lines up from the \u03b4-disk to make a disk on the function, which is called the image of delta disk. Then there are dashed lines from this disk to the circle around the point L, which is called the \u0949-neighborhood of L.\" width=\"646\" height=\"581\" \/><\/p>\n<p id=\"caption-attachment-971\" class=\"wp-caption-text\">Figure 2.\u00a0The limit of a function involving two variables requires that\u00a0[latex]f(x,y)[\/latex] be within\u00a0[latex]\\varepsilon[\/latex] of\u00a0[latex]L[\/latex]\u00a0whenever [latex](x,y)[\/latex]\u00a0is within [latex]\\delta[\/latex] of\u00a0[latex](a,b)[\/latex]. The smaller the value of\u00a0[latex]\\varepsilon[\/latex],\u00a0the smaller the value of\u00a0[latex]\\delta[\/latex].<\/p>\n<\/div>\n<p>Proving that a limit exists using the definition of a <span id=\"term158\" class=\"no-emphasis\" data-type=\"term\">limit of a function of two variables<\/span> can be challenging. Instead, we use the following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of the formulas in the limit laws theorem in <a href=\"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/introduction-to-the-limit-laws\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"8b89d172-2927-466f-8661-01abc7ccdba4\" data-page-slug=\"2-3-the-limit-laws\">The Limit Laws<\/a>.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Limit LAws for Functions of Two Variables Theorem<\/h3>\n<hr \/>\n<p>Let [latex]f\\,(x,\\ y)[\/latex]\u00a0and [latex]g\\,(x,\\ y)[\/latex] be defined for all [latex](x,\\ y)\\neq(a,\\ b)[\/latex] in a neighborhood around [latex](a,\\ b)[\/latex], and assume\u00a0the neighborhood is contained completely inside the domain of [latex]f[\/latex]. Assume that [latex]L[\/latex] and [latex]M[\/latex] are real numbers such that [latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}f\\,(x,\\ y)=L[\/latex] and [latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}g\\,(x,\\ y)=M[\/latex], and let [latex]c[\/latex] be a constant.\u00a0Then each of the following statements holds:<\/p>\n<p><strong>Constant\u00a0Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}c=c[\/latex]<\/p>\n<p><strong>Identity Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}x} & =\\hfill & {a} \\hfill \\\\ \\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}y} & =\\hfill & {b} \\end{array}[\/latex]<\/p>\n<p><strong>Sum Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)+g\\,(x,\\ y))=L+M[\/latex]<\/p>\n<p><strong>Difference Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)-(g\\,(x,\\ y))=L-M[\/latex]<\/p>\n<p><strong>Constant Multiple<\/strong> <strong>Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(cf\\,(x,\\ y))=cL[\/latex]<\/p>\n<p><strong>Product Law:\u00a0<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y)\\,g\\,(x,\\ y))=LM[\/latex]<\/p>\n<p><strong>Quotient Law:\u00a0<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\frac{f\\,(x,\\ y)}{g\\,(x,\\ y)}=\\frac{L}{M}[\/latex] for [latex]M\\neq{0}[\/latex]<\/p>\n<p><strong>Power Law:\u00a0<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}(f\\,(x,\\ y))^{n}=L^{n}[\/latex]<\/p>\n<p>for any positive integer [latex]n[\/latex].<\/p>\n<p><strong>Root Law:<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\sqrt[n]{f\\,(x,\\ y)}=\\sqrt[n]{L}[\/latex]<\/p>\n<p>for all [latex]L[\/latex] if [latex]n[\/latex] is odd and positive, and for [latex]L\\,\\geq\\,0[\/latex] if [latex]n[\/latex] is even and positive.<\/p>\n<\/div>\n<p>The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws to finding limits of various functions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Limit of a Function of Two Variables<\/h3>\n<p>Find each of the following limits:<\/p>\n<ol>\n<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[\/latex]<\/li>\n<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}\\frac{2x+3y}{4x-3y}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>First use the sum and difference laws to separate the terms:<\/li>\n<\/ol>\n<p>[latex]\\hspace{5cm}\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\hspace{5cm}=\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x^{2}\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}2xy\\bigg) +\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y^{2}\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}4x\\bigg) \\\\ \\hspace{5cm}+\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Next,\u00a0use the constant multiple law on the second, third, fourth, and fifth limits:<\/p>\n<div style=\"text-align: center;\">[latex]\\hspace{5cm}=\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x^{2}\\bigg)-2\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}xy\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y^{2}\\bigg)-4\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg) \\\\ \\hspace{5cm} +3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Now, use the power law on the first and third limits, and the product law on the second limit:<\/p>\n<div style=\"text-align: center;\">[latex]\\hspace{5cm} =\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg)^{2}-2\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}xy\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)^{2} \\\\ \\hspace{5cm} -4\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\bigg)+3\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\bigg)-\\bigg(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}6\\bigg).[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Last, use the identity laws on the first six limits and the constant law on the last limit:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)}& =\\hfill & {(2)^{2}-2(2)(-1)+3(-1)^{2}-4(2)+3(-1)-6} \\hfill \\\\ \\hfill & =\\hfill & {-6} \\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<ol>\n<li>Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(4x-3y)} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}4x-\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y}\\hfill \\\\ \\hfill & =\\hfill & {4\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\big)-3\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\big)} \\hfill \\\\ \\hfill & =\\hfill & {4(2)-3(-1)=11.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(2x+3y)} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}2x+\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}3y}\\hfill \\\\ \\hfill & =\\hfill & {2\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}x\\big)+3\\big(\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}y\\big)} \\hfill \\\\ \\hfill & =\\hfill & {2(2)+3(-1)}\\hfill \\\\ \\hfill & =\\hfill & {1.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Therefore, according to the quotient law we have<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}\\frac{2x+3y}{4x-3y}=\\frac{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(2x+3y)}{\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ -1)}(4x-3y)}=\\frac{1}{11}.[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>Evaluate the following limit:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(5,\\ -2)}\\sqrt[3]{\\frac{x^{2}-y}{y^{2}+x-1}.}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1367793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1367793933114\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(5,\\ -2)}\\sqrt[3]{\\frac{x^{2}-y}{y^{2}+x-1}}=\\frac{3}{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Since we are taking the limit of a function of two variables, the point [latex](a,\\ b)[\/latex] is in [latex]\\mathbb{R}^{2}[\/latex],\u00a0and it is possible to approach this point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the path taken toward\u00a0[latex](a,\\ b)[\/latex].\u00a0If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of path taken.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Limits that Fail to Exist<\/h3>\n<p>Show that neither of the following limits exist:<\/p>\n<ol>\n<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{2xy}{3x^{2}+y^{2}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1197793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1197793933114\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The domain of the function [latex]f\\,(x,\\ y)=\\frac{2xy}{3x^{2}+y^{2}}[\/latex]\u00a0consists of all points in the [latex]xy[\/latex]-plane\u00a0except for the point [latex](0,\\ 0)[\/latex]\u00a0(Figure 3). To show that the limit does not exist as [latex](x,\\ y)[\/latex] approaches [latex](0,\\ 0)[\/latex],\u00a0we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point [latex](0,\\ 0)[\/latex].\u00a0First, consider the line [latex]y=0[\/latex] in the [latex]xy[\/latex]-plane. Substituting\u00a0[latex]y=0[\/latex] into [latex]f\\,(x,\\ y)[\/latex] gives\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ 0)=\\frac{2x(0)}{3x^{2}+0^{2}}=0}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>For any value of [latex]x[\/latex].\u00a0Therefore the value of [latex]f[\/latex]\u00a0remains constant for any point on the [latex]x[\/latex]-axis, and as [latex]y[\/latex] approaches zero, the function remains fixed at zero.\u00a0Next, consider the line [latex]y=x[\/latex]. Substituting [latex]y=x[\/latex] into [latex]f\\,(x,\\ y)[\/latex] gives<\/p>\n<div style=\"text-align: center;\">[latex]\\large{f\\,(x,\\ x)=\\frac{2x(x)}{3x^{2}+x^{2}}=\\frac{2x^{2}}{4x^{2}}=\\frac{1}{2}}.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>This is true for any point on the line [latex]y=x[\/latex]. If we let [latex]x[\/latex] approach zero\u00a0while staying on this line, the value of the function remains fixed at [latex]\\frac{1}{2}[\/latex],\u00a0regardless of how small [latex]x[\/latex] is.\u00a0Choose a value for [latex]\\epsilon[\/latex] that is less than [latex]1\/2-[\/latex]say, [latex]1\/4[\/latex]. Then, no matter how small a [latex]\\delta[\/latex] disk\u00a0we draw around [latex](0,\\ 0)[\/latex], the values of [latex]f\\,(x,\\ y)[\/latex] for points inside that [latex]\\delta[\/latex] disk will include both [latex]0[\/latex] and [latex]\\frac{1}{2}.[\/latex] Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.<\/p>\n<div id=\"attachment_973\" style=\"width: 461px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-973\" class=\"size-full wp-image-973\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28224845\/4-2-3.jpeg\" alt=\"In xyz space, the function f(x, y) = 2xy\/(3x2 + y2) is shown, which is a slightly twisted plane, with values of 0 along the line y = 0 and values of \u00bd along the line y = x.\" width=\"451\" height=\"288\" \/><\/p>\n<p id=\"caption-attachment-973\" class=\"wp-caption-text\">Figure 3. Graph of the function [latex]\\small{f(x,y)=(2xy)\/(3x^{2}+y^{2})}[\/latex]. Along the line [latex]\\small{y=0}[\/latex], the function is equal to zero; along the line [latex]\\small{y=x}[\/latex], the function is equal to [latex]\\small{\\frac{1}{2}}[\/latex].<\/p>\n<\/div>\n<\/li>\n<li>In a similar fashion to 1., we can approach the origin along any straight line passing through the origin. If we try the [latex]x[\/latex]-axis (i.e., [latex]y=0[\/latex]),\u00a0then the function remains fixed at zero. The same is true for the [latex]y[\/latex]-axis.\u00a0Suppose we approach the origin along a straight line of slope [latex]k[\/latex]\u00a0The equation of this line is [latex]y=kx[\/latex]. Then the limit becomes\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4x(kx)^{2}}{x^{2}+3(kx)^{4}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4k^{2}x^{3}}{x^{2}+3k^{4}x^{4}}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4k^{2}x}{1+3k^{4}x^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {\\frac{\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}4k^{2}x}{\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}(1+3k^{4}x^{2})}}\\hfill \\\\ \\hfill & =\\hfill & {0}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Regardless of the value of [latex]k[\/latex].\u00a0It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation [latex]x=y^{2}[\/latex]. Substituting [latex]y^{2}[\/latex] in place of [latex]x[\/latex] in [latex]f\\,(x,\\ y)[\/latex] gives<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}} & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4(y^{2})y^{2}}{(y^{2})^{2}+3y^{4}}}\\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4y^{4}}{y^{4}+3y^{4}}} \\hfill \\\\ \\hfill & =\\hfill & {\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}1}\\hfill \\\\ \\hfill & =\\hfill & {1.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>By the same logic\u00a0in 1., it is impossible to find a [latex]\\delta[\/latex] disk\u00a0around the origin that satisfies the definition of the limit for any value of [latex]\\epsilon\\,<\\,1[\/latex]. Therefore, [latex]\\displaystyle\\lim_{(x,\\ y)\\to(0,\\ 0)}\\frac{4xy^{2}}{x^{2}+3y^{4}}[\/latex] does not exist.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>Show that<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ 1)}\\frac{(x-2)(y-1)}{(x-2)^{2}+(y-1)^{2}}[\/latex]<\/p>\n<p>does not exist.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1667793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1667793933114\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">If [latex]y=k\\,(x-2)+1[\/latex], then [latex]\\displaystyle\\lim_{(x,\\ y)\\to(2,\\ 1)}\\frac{(x-2)(y-1)}{(x-2)^{2}+(y-1)^{2}}=\\frac{k}{1+k^{2}}[\/latex]. Since the answer depends on [latex]k[\/latex], the limit fails to exist.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186150&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=pWvDsHUhTfY&amp;video_target=tpm-plugin-dfjtf1gg-pWvDsHUhTfY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.7_transcript.html\">transcript for \u201cCP 4.7\u201d here (opens in new window).<\/a><\/div>\n<h2>Interior Points and Boundary Points<\/h2>\n<p>To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]S[\/latex] be a subset of [latex]\\mathbb{R}^{2}[\/latex]\u00a0(Figure 4).<\/p>\n<ol>\n<li>A point [latex]P_0[\/latex] is called an\u00a0<strong>interior point<\/strong> of [latex]S[\/latex] if there is a [latex]\\delta[\/latex] disk centered around [latex]P_0[\/latex] contained completely in [latex]S[\/latex].<\/li>\n<li>A point [latex]P_0[\/latex] is called a\u00a0<strong>boundary point\u00a0<\/strong>of [latex]S[\/latex] if every [latex]\\delta[\/latex] disk centered around [latex]P_0[\/latex]\u00a0contains points both inside and outside [latex]S[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"attachment_974\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-974\" class=\"size-full wp-image-974\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225116\/4-2-4.jpeg\" alt=\"On the xy plane, a closed shape is drawn. There is a point (\u20131, 1) drawn on the inside of the shape, and there is a point (2, 3) drawn on the boundary. Both of these points are the centers of small circles.\" width=\"417\" height=\"422\" \/><\/p>\n<p id=\"caption-attachment-974\" class=\"wp-caption-text\">Figure 4.\u00a0In the set [latex]S[\/latex] shown,\u00a0[latex]\\small{(-1,1)}[\/latex] is an interior point and\u00a0[latex]\\small{(2,3)}[\/latex]\u00a0is a boundary point.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]S[\/latex] be a subset of [latex]\\mathbb{R}^{2}[\/latex]\u00a0(Figure 4).<\/p>\n<ol>\n<li>[latex]S[\/latex] is called an\u00a0<strong>open set<\/strong> if every point of [latex]S[\/latex] is an interior point.<\/li>\n<li>[latex]S[\/latex]\u00a0is called a <strong><span id=\"term162\" data-type=\"term\">closed set<\/span><\/strong> if it contains all its boundary points.<\/li>\n<\/ol>\n<\/div>\n<p>An example of an open set is a [latex]\\delta[\/latex]\u00a0disk. If we include the boundary of the disk, then it becomes a closed set. A set that contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary of a [latex]\\delta[\/latex]\u00a0disk but not the other half, then the set is neither open nor closed.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]S[\/latex] be a subset of [latex]\\mathbb{R}^{2}[\/latex]\u00a0(Figure 4).<\/p>\n<ol>\n<li>An open set [latex]S[\/latex]\u00a0is a <strong><span id=\"term163\" data-type=\"term\">connected set<\/span> <\/strong>if it cannot be represented as the union of two or more disjoint, nonempty open subsets.<\/li>\n<li>A set [latex]S[\/latex]\u00a0is a <strong><span id=\"term164\" data-type=\"term\">region<\/span><\/strong> if it is open, connected, and nonempty.<\/li>\n<\/ol>\n<\/div>\n<p>The definition of a limit of a function of two variables requires the [latex]\\delta[\/latex]\u00a0disk to be contained inside the domain of the function. However, if we wish to find the limit of a function at a boundary point of the domain, the [latex]\\delta[\/latex] disk is not contained inside the domain. By definition, some of the points of the\u00a0[latex]\\delta[\/latex] disk\u00a0are inside the domain and some are outside. Therefore, we need only consider points that are inside both the [latex]\\delta[\/latex]\u00a0disk and the domain of the function. This leads to the definition of the limit of a function at a boundary point.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]f[\/latex] be a function of two variables, [latex]x[\/latex] and [latex]y[\/latex], and suppose [latex](a,\\ b)[\/latex]\u00a0is on the boundary of the domain of [latex]f[\/latex]. Then, the limit of [latex]f\\,(x,\\ y)[\/latex] as [latex](x,\\ y)[\/latex] approaches [latex](a,\\ b)[\/latex] is [latex]L[\/latex], written<\/p>\n<div style=\"text-align: center;\">[latex]\\large{\\displaystyle\\lim_{(x,\\ y)\\to(a,\\ b)}\\,f\\,(x,\\ y)=L}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p>if for any [latex]\\epsilon\\,>\\,0[\/latex], there exists a number [latex]\\delta\\,>\\,0[\/latex] such that for any point [latex](x,\\ y)[\/latex] inside the domain of [latex]f[\/latex] and within a suitably small distance positive [latex]\\delta[\/latex] of [latex](a,\\ b)[\/latex], the value of [latex]f\\,(x,\\ y)[\/latex] is no more than [latex]\\epsilon[\/latex] away from [latex]L[\/latex]\u00a0(Figure 2). Using symbols, we can write: For any [latex]\\epsilon\\,>\\,0[\/latex], there exists a number [latex]\\delta\\,>\\,0[\/latex] such that<\/p>\n<div style=\"text-align: center;\">[latex]\\large{|f\\,(x,\\ y)-L|\\,<\\,\\epsilon}[\/latex] whenever [latex]\\large{0\\,<\\,\\sqrt{(x-a)^{2}+(y-b)^{2}}\\,<\\,\\delta}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Limit of a Function at a Boundary Point<\/h3>\n<p>Prove [latex]\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}\\sqrt{25-x^{2}-y^{2}}=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1667743933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1667743933114\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: left;\">The domain of the function [latex]f\\,(x,\\ y)=\\sqrt{25-x^{2}-y^{2}}[\/latex] is [latex]\\{(x,\\ y)\\in\\mathbb{R}^{2}\\,|\\,x^{2}+y^{2}\\,\\leq\\,25\\}[\/latex], which\u00a0is a circle of radius [latex]5[\/latex]\u00a0centered at the origin, along with its interior as shown in the following graph.<\/div>\n<div>\n<div id=\"attachment_976\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-976\" class=\"wp-image-976 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225404\/4-2-5.jpeg\" alt=\"A circle with radius 5 centered at the origin with its interior filled in.\" width=\"417\" height=\"422\" \/><\/p>\n<p id=\"caption-attachment-976\" class=\"wp-caption-text\">Figure 5. Domain of the function [latex]\\small{f(x,y)=\\sqrt{25-x^{2}-y^{2}}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<div>We can\u00a0use the limit laws, which apply to limits at the boundary of domains as well as interior points:<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}\\sqrt{25-x^{2}-y^{2}}} & =\\hfill & {\\sqrt{\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}(25-x^{2}-y^{2})}} \\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}25-\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}x^{2}-\\displaystyle\\lim_{(x,\\ y)\\to(4,\\ 3)}y^{2}}} \\hfill \\\\ \\hfill & =\\hfill & {\\sqrt{25-4^{2}-3^{2}}}\\hfill \\\\ \\hfill & =\\hfill & {0.}\\hfill \\\\ \\hfill \\end{array}[\/latex]<\/div>\n<div>See the following graph.<\/div>\n<div>\n<div id=\"attachment_977\" style=\"width: 363px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-977\" class=\"size-full wp-image-977\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/28225533\/4-2-6.jpeg\" alt=\"The upper hemisphere in xyz space with radius 5 and center the origin.\" width=\"353\" height=\"378\" \/><\/p>\n<p id=\"caption-attachment-977\" class=\"wp-caption-text\">Figure 6.\u00a0Graph of the function [latex]\\small{f(x,y)=\\sqrt{25-x^{2}-y^{2}}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>Evaluate the following limit:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(5,-2)}\\sqrt{29-x^{2}-y^{2}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1667793733114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1667793733114\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\displaystyle\\lim_{(x,\\ y)\\to(5,-2)}\\sqrt{29-x^{2}-y^{2}}=0[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186151&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=zYJIT-NNVKM&amp;video_target=tpm-plugin-0y6dbuw0-zYJIT-NNVKM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.8_transcript.html\">transcript for \u201cCP 4.8\u201d here (opens in new window).<\/a><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3912\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 4.7. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>CP 4.8. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 4.7\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 4.8\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3912","chapter","type-chapter","status-publish","hentry"],"part":22,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3912","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3912\/revisions"}],"predecessor-version":[{"id":5900,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3912\/revisions\/5900"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/22"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3912\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3912"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3912"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3912"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3912"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}