{"id":2286,"date":"2021-09-21T15:11:50","date_gmt":"2021-09-21T15:11:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-alternating-series\/"},"modified":"2024-01-03T15:30:46","modified_gmt":"2024-01-03T15:30:46","slug":"skills-review-for-alternating-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-alternating-series\/","title":{"raw":"Skills Review for Alternating Series and Ratio and Root Tests","rendered":"Skills Review for Alternating Series and Ratio and Root Tests"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Apply factorial notation<\/li>\r\n \t<li>Simplify expressions using the Product Property of Exponents<\/li>\r\n \t<li>Calculate the limit of a function as \ud835\udc65 increases or decreases without bound<\/li>\r\n \t<li>Recognize when to apply L\u2019H\u00f4pital\u2019s rule<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Alternating Series and Ratio and Root Tests sections, we will learn about the last few methods that can be used to determine whether an infinite series diverges or converges. Here we will review how to use factorial notation, use product rule for exponents, take limits at infinity, and apply\u00a0L'Hopital's Rule.\r\n<h2>Apply Factorial Notation<\/h2>\r\nRecall that\u00a0<strong>[latex]n[\/latex] factorial<\/strong>, written as [latex]n![\/latex], is the product of the positive integers from 1 to [latex]n[\/latex]. For example,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}4!&amp;=4\\cdot 3\\cdot 2\\cdot 1=24 \\\\ 5!&amp;=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120\\\\ \\text{ } \\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\nAn example of formula containing a <strong>factorial<\/strong> is [latex]{a}_{n}=\\left(n+1\\right)![\/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{6}=\\left(6+1\\right)!=7!=7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=5040 \\\\ \\text{ }\\end{align}[\/latex]<\/div>\r\n<div><\/div>\r\nThe factorial of any whole number [latex]n[\/latex] is [latex]n\\left(n - 1\\right)![\/latex] We can therefore also think of [latex]5![\/latex] as [latex]5\\cdot 4!\\text{.}[\/latex]\r\n<div class=\"textbox\">\r\n<h3>A GENERAL NOTE: FACTORIAL<\/h3>\r\n<strong><em>n<\/em> factorial<\/strong> is a mathematical operation that can be defined using a recursive formula. The factorial of [latex]n[\/latex], denoted [latex]n![\/latex], is defined for a positive integer [latex]n[\/latex] as:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}0!=1\\\\ 1!=1\\\\ n!=n\\left(n - 1\\right)\\left(n - 2\\right)\\cdots \\left(2\\right)\\left(1\\right)\\text{, for }n\\ge 2\\end{array}[\/latex]<\/div>\r\nThe special case [latex]0![\/latex] is defined as [latex]0!=1[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=61071&amp;amp;theme=oea&amp;amp;iframe_resize_id=mom2[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nExpand [latex](n+3)![\/latex].\r\n[reveal-answer q=\"953701\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953701\"]\r\n\r\n[latex](n+3)!=(n+3)(n+2)(n+1)n(n-1)\\cdot...\\cdot1[\/latex]\r\n\r\nNotice also that [latex](n+3)!=(n+3)(n+2)(n+1)n![\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use the Product Rule for Exponents<\/h2>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Product Rule of Exponents<\/h3>\r\nFor any real number [latex]a[\/latex] and natural numbers [latex]m[\/latex] and [latex]n[\/latex], the product rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Product Rule<\/h3>\r\nWrite each of the following products with a single base. Do not simplify further.\r\n<ol>\r\n \t<li>[latex]{t}^{5}\\cdot {t}^{3}[\/latex]<\/li>\r\n \t<li>[latex]\\left(-3\\right)^{5}\\cdot \\left(-3\\right)[\/latex]<\/li>\r\n \t<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"878162\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"878162\"]\r\nUse the product rule to simplify each expression.\r\n<ol>\r\n \t<li>[latex]{t}^{5}\\cdot {t}^{3}={t}^{5+3}={t}^{8}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(-3\\right)}^{5}\\cdot \\left(-3\\right)={\\left(-3\\right)}^{5}\\cdot {\\left(-3\\right)}^{1}={\\left(-3\\right)}^{5+1}={\\left(-3\\right)}^{6}[\/latex]<\/li>\r\n \t<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\r\n<\/ol>\r\nAt first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}=\\left({x}^{2}\\cdot {x}^{5}\\right)\\cdot {x}^{3}=\\left({x}^{2+5}\\right)\\cdot {x}^{3}={x}^{7}\\cdot {x}^{3}={x}^{7+3}={x}^{10}[\/latex]<\/div>\r\nNotice we get the same result by adding the three exponents in one step.\r\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1961&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox\"><h3>RECALL<\/h3>\r\nFor any real number [latex]a[\/latex] and positive integers [latex]m[\/latex] and [latex]n[\/latex], the power rule of exponents states that\r\n<center>[latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/center> \u00a0\r\n\r\nFor an expression like\u00a0[latex](a^2)^3[\/latex], you have a base of\u00a0[latex]a[\/latex] raised to the power of [latex]2[\/latex], which is then raised to another power of [latex]3[\/latex]. Multiply the exponents\u00a0[latex]2[\/latex]\u00a0and\u00a0[latex]3[\/latex] to find the new exponent for [latex]a[\/latex]. This gives you\u00a0[latex]a^{2\\cdot3}[\/latex] or\u00a0[latex]a^6[\/latex]. Always remember: when an exponent is raised to another exponent, multiply the exponents to simplify the expression.<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify the expression [latex](3a^2b)^3 \\cdot (2ab^4)[\/latex].\r\n[reveal-answer q=\"953721\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953721\"]\r\n\r\n[latex](xy)^n=x^n \\cdot y^n[\/latex], so we apply the exponent [latex]3[\/latex] to both [latex]3a^2[\/latex] and [latex]b[\/latex] within the parentheses.\r\n\r\n<center>[latex](3a^2b)^3 = 3^3 \\cdot (a^2)^3 \\cdot b^3[\/latex]<\/center>\r\nNow we'll compute the powers individually.\r\n\r\n<center>[latex]\\begin{array}{l}\r\n3^3 = 27 \\\\\r\n(a^2)^3 = a^{2\\cdot3} = a^6 \\\\\r\nb^3 = b^3 \\text{ (no change since it's already in exponent form)}\r\n\\end{array}[\/latex]<\/center>\r\nPut together the results of the individual powers.\r\n\r\n<center>[latex](3a^2b)^3=27 \\cdot a^6 \\cdot b^3[\/latex]<\/center>\r\nNow we multiply the result by the second term [latex]2ab^4[\/latex]. Since we are dealing with multiplication, we can multiply coefficients (the numerical parts) and variables with the same base separately.\r\n\r\n<center>[latex](27\u22c5a^6 \\cdot b^3) \\cdot (2ab^4) = (27 \\cdot 2) \\cdot (a^6 \\cdot a) \\cdot (b^3 \\cdot b^4)[\/latex]<\/center>\r\nMultiply the coefficients and add the exponents for the variables with the same base.\r\n\r\n<center>[latex]\\begin{array}{l}\r\n27 \\cdot 2 = 54 \\\\\r\na^6 \\cdot a = a^{6+1} = a^7 \\\\\r\nb^3 \\cdot b^4 = b^{3+4} = b^7\r\n\\end{array}[\/latex]<\/center>\r\nFinally, combine all the simplified terms to get the final answer.\r\n\r\n<center>[latex]54 \\cdot a^7 \\cdot b^7[\/latex]<\/center>\r\nSo, the expression [latex](3a^2b)^3 \\cdot (2ab^4)[\/latex] simplifies to [latex]54a^7b^7[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSimplify the expression [latex](2y^2)^3 \\cdot (4y^5).[\/latex]\r\n[reveal-answer q=\"953799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"953799\"]\r\n\r\nFirst, apply the power to the first term, [latex](2y^2)^3[\/latex], by multiplying the exponent outside the parenthesis with each exponent inside the parenthesis:\r\n\r\n<center>[latex](2y^2)^3=2^3 \\cdot (y^2)^3=8y^6[\/latex]<\/center>\r\nNow, multiply this result by [latex]4y^5[\/latex]:\r\n\r\n<center>[latex]8y^6 \\cdot 4y^5 = (8 \\cdot 4) \\cdot y^{6+5} = 32y^{11}[\/latex]<\/center>\r\nTherefore, the expression simplifies to [latex]32y^{11}[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Take Limits at Infinity<\/h2>\r\n<em><strong>(see <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-sequences\/\" target=\"_blank\" rel=\"noopener\">Module 5, Skills Review for Sequences<\/a>.)<\/strong><\/em>\r\n<h2>Infinite Limits at Infinity<\/h2>\r\n<em><strong>(see <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-sequences\/\" target=\"_blank\" rel=\"noopener\">Module 5, Skills Review for Sequences<\/a>.)<\/strong><\/em>\r\n<h2>Apply L'H\u00f4pital's Rule<\/h2>\r\n<em><strong>(see <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-sequences\/\" target=\"_blank\" rel=\"noopener\">Module 5, Skills Review for Sequences<\/a>.)<\/strong><\/em>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Apply factorial notation<\/li>\n<li>Simplify expressions using the Product Property of Exponents<\/li>\n<li>Calculate the limit of a function as \ud835\udc65 increases or decreases without bound<\/li>\n<li>Recognize when to apply L\u2019H\u00f4pital\u2019s rule<\/li>\n<\/ul>\n<\/div>\n<p>In the Alternating Series and Ratio and Root Tests sections, we will learn about the last few methods that can be used to determine whether an infinite series diverges or converges. Here we will review how to use factorial notation, use product rule for exponents, take limits at infinity, and apply\u00a0L&#8217;Hopital&#8217;s Rule.<\/p>\n<h2>Apply Factorial Notation<\/h2>\n<p>Recall that\u00a0<strong>[latex]n[\/latex] factorial<\/strong>, written as [latex]n![\/latex], is the product of the positive integers from 1 to [latex]n[\/latex]. For example,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}4!&=4\\cdot 3\\cdot 2\\cdot 1=24 \\\\ 5!&=5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=120\\\\ \\text{ } \\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>An example of formula containing a <strong>factorial<\/strong> is [latex]{a}_{n}=\\left(n+1\\right)![\/latex]. The sixth term of the sequence can be found by substituting 6 for [latex]n[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{6}=\\left(6+1\\right)!=7!=7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2\\cdot 1=5040 \\\\ \\text{ }\\end{align}[\/latex]<\/div>\n<div><\/div>\n<p>The factorial of any whole number [latex]n[\/latex] is [latex]n\\left(n - 1\\right)![\/latex] We can therefore also think of [latex]5![\/latex] as [latex]5\\cdot 4!\\text{.}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A GENERAL NOTE: FACTORIAL<\/h3>\n<p><strong><em>n<\/em> factorial<\/strong> is a mathematical operation that can be defined using a recursive formula. The factorial of [latex]n[\/latex], denoted [latex]n![\/latex], is defined for a positive integer [latex]n[\/latex] as:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}0!=1\\\\ 1!=1\\\\ n!=n\\left(n - 1\\right)\\left(n - 2\\right)\\cdots \\left(2\\right)\\left(1\\right)\\text{, for }n\\ge 2\\end{array}[\/latex]<\/div>\n<p>The special case [latex]0![\/latex] is defined as [latex]0!=1[\/latex].<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm61071\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=61071&#38;theme=oea&#38;iframe_resize_id=ohm61071&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Expand [latex](n+3)![\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953701\">Show Solution<\/span><\/p>\n<div id=\"q953701\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](n+3)!=(n+3)(n+2)(n+1)n(n-1)\\cdot...\\cdot1[\/latex]<\/p>\n<p>Notice also that [latex](n+3)!=(n+3)(n+2)(n+1)n![\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Use the Product Rule for Exponents<\/h2>\n<div class=\"textbox\">\n<h3>A General Note: The Product Rule of Exponents<\/h3>\n<p>For any real number [latex]a[\/latex] and natural numbers [latex]m[\/latex] and [latex]n[\/latex], the product rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Product Rule<\/h3>\n<p>Write each of the following products with a single base. Do not simplify further.<\/p>\n<ol>\n<li>[latex]{t}^{5}\\cdot {t}^{3}[\/latex]<\/li>\n<li>[latex]\\left(-3\\right)^{5}\\cdot \\left(-3\\right)[\/latex]<\/li>\n<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q878162\">Show Solution<\/span><\/p>\n<div id=\"q878162\" class=\"hidden-answer\" style=\"display: none\">\nUse the product rule to simplify each expression.<\/p>\n<ol>\n<li>[latex]{t}^{5}\\cdot {t}^{3}={t}^{5+3}={t}^{8}[\/latex]<\/li>\n<li>[latex]{\\left(-3\\right)}^{5}\\cdot \\left(-3\\right)={\\left(-3\\right)}^{5}\\cdot {\\left(-3\\right)}^{1}={\\left(-3\\right)}^{5+1}={\\left(-3\\right)}^{6}[\/latex]<\/li>\n<li>[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}[\/latex]<\/li>\n<\/ol>\n<p>At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first two.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}=\\left({x}^{2}\\cdot {x}^{5}\\right)\\cdot {x}^{3}=\\left({x}^{2+5}\\right)\\cdot {x}^{3}={x}^{7}\\cdot {x}^{3}={x}^{7+3}={x}^{10}[\/latex]<\/div>\n<p>Notice we get the same result by adding the three exponents in one step.<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}\\cdot {x}^{5}\\cdot {x}^{3}={x}^{2+5+3}={x}^{10}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm1961\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1961&#38;theme=oea&#38;iframe_resize_id=ohm1961&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>RECALL<\/h3>\n<p>For any real number [latex]a[\/latex] and positive integers [latex]m[\/latex] and [latex]n[\/latex], the power rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/div>\n<p> \u00a0<\/p>\n<p>For an expression like\u00a0[latex](a^2)^3[\/latex], you have a base of\u00a0[latex]a[\/latex] raised to the power of [latex]2[\/latex], which is then raised to another power of [latex]3[\/latex]. Multiply the exponents\u00a0[latex]2[\/latex]\u00a0and\u00a0[latex]3[\/latex] to find the new exponent for [latex]a[\/latex]. This gives you\u00a0[latex]a^{2\\cdot3}[\/latex] or\u00a0[latex]a^6[\/latex]. Always remember: when an exponent is raised to another exponent, multiply the exponents to simplify the expression.<\/p><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify the expression [latex](3a^2b)^3 \\cdot (2ab^4)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953721\">Show Solution<\/span><\/p>\n<div id=\"q953721\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](xy)^n=x^n \\cdot y^n[\/latex], so we apply the exponent [latex]3[\/latex] to both [latex]3a^2[\/latex] and [latex]b[\/latex] within the parentheses.<\/p>\n<div style=\"text-align: center;\">[latex](3a^2b)^3 = 3^3 \\cdot (a^2)^3 \\cdot b^3[\/latex]<\/div>\n<p>Now we&#8217;ll compute the powers individually.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}  3^3 = 27 \\\\  (a^2)^3 = a^{2\\cdot3} = a^6 \\\\  b^3 = b^3 \\text{ (no change since it's already in exponent form)}  \\end{array}[\/latex]<\/div>\n<p>Put together the results of the individual powers.<\/p>\n<div style=\"text-align: center;\">[latex](3a^2b)^3=27 \\cdot a^6 \\cdot b^3[\/latex]<\/div>\n<p>Now we multiply the result by the second term [latex]2ab^4[\/latex]. Since we are dealing with multiplication, we can multiply coefficients (the numerical parts) and variables with the same base separately.<\/p>\n<div style=\"text-align: center;\">[latex](27\u22c5a^6 \\cdot b^3) \\cdot (2ab^4) = (27 \\cdot 2) \\cdot (a^6 \\cdot a) \\cdot (b^3 \\cdot b^4)[\/latex]<\/div>\n<p>Multiply the coefficients and add the exponents for the variables with the same base.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}  27 \\cdot 2 = 54 \\\\  a^6 \\cdot a = a^{6+1} = a^7 \\\\  b^3 \\cdot b^4 = b^{3+4} = b^7  \\end{array}[\/latex]<\/div>\n<p>Finally, combine all the simplified terms to get the final answer.<\/p>\n<div style=\"text-align: center;\">[latex]54 \\cdot a^7 \\cdot b^7[\/latex]<\/div>\n<p>So, the expression [latex](3a^2b)^3 \\cdot (2ab^4)[\/latex] simplifies to [latex]54a^7b^7[\/latex].\n<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Simplify the expression [latex](2y^2)^3 \\cdot (4y^5).[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q953799\">Show Solution<\/span><\/p>\n<div id=\"q953799\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, apply the power to the first term, [latex](2y^2)^3[\/latex], by multiplying the exponent outside the parenthesis with each exponent inside the parenthesis:<\/p>\n<div style=\"text-align: center;\">[latex](2y^2)^3=2^3 \\cdot (y^2)^3=8y^6[\/latex]<\/div>\n<p>Now, multiply this result by [latex]4y^5[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]8y^6 \\cdot 4y^5 = (8 \\cdot 4) \\cdot y^{6+5} = 32y^{11}[\/latex]<\/div>\n<p>Therefore, the expression simplifies to [latex]32y^{11}[\/latex].\n<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Take Limits at Infinity<\/h2>\n<p><em><strong>(see <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-sequences\/\" target=\"_blank\" rel=\"noopener\">Module 5, Skills Review for Sequences<\/a>.)<\/strong><\/em><\/p>\n<h2>Infinite Limits at Infinity<\/h2>\n<p><em><strong>(see <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-sequences\/\" target=\"_blank\" rel=\"noopener\">Module 5, Skills Review for Sequences<\/a>.)<\/strong><\/em><\/p>\n<h2>Apply L&#8217;H\u00f4pital&#8217;s Rule<\/h2>\n<p><em><strong>(see <a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/skills-review-for-sequences\/\" target=\"_blank\" rel=\"noopener\">Module 5, Skills Review for Sequences<\/a>.)<\/strong><\/em><\/p>\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-2286\">\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>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <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>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <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>\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":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen 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