{"id":457,"date":"2021-02-04T15:27:58","date_gmt":"2021-02-04T15:27:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&p=457"},"modified":"2021-06-22T05:39:39","modified_gmt":"2021-06-22T05:39:39","slug":"problem-set-the-limit-laws","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/problem-set-the-limit-laws\/","title":{"raw":"Problem Set: The Limit Laws","rendered":"Problem Set: The Limit Laws"},"content":{"raw":"

In the following exercises (1-4), use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\lim}(4x^2-2x+3)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572597974\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572597974\"]\r\n

Use constant multiple law and difference law: [latex]\\underset{x\\to 0}{\\lim}(4x^2-2x+3)=4\\underset{x\\to 0}{\\lim}x^2-2\\underset{x\\to 0}{\\lim}x+\\underset{x\\to 0}{\\lim}3=3[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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2.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}\\frac{x^3+3x^2+5}{4-7x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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3.\u00a0<\/strong>[latex]\\underset{x\\to -2}{\\lim}\\sqrt{x^2-6x+3}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571574734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571574734\"]\r\n

Use root law: [latex]\\underset{x\\to -2}{\\lim}\\sqrt{x^2-6x+3}=\\sqrt{\\underset{x\\to -2}{\\lim}(x^2-6x+3)}=\\sqrt{19}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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\r\n\r\n4.\u00a0<\/strong>[latex]\\underset{x\\to -1}{\\lim}(9x+1)^2[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (5-10), use direct substitution to evaluate each limit.<\/p>\r\n\r\n

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5.\u00a0<\/strong>[latex]\\underset{x\\to 7}{\\lim}x^2[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571654822\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571654822\"]\r\n

49<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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6.\u00a0<\/strong>[latex]\\underset{x\\to -2}{\\lim}(4x^2-1)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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7.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{1}{1+ \\sin x}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571654916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571654916\"]\r\n

1<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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8.\u00a0<\/strong>[latex]\\underset{x\\to 2}{\\lim}e^{2x-x^2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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9.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{2-7x}{x+6}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572482577\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572482577\"]\r\n

[latex]-\\frac{5}{7}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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10.\u00a0<\/strong>[latex]\\underset{x\\to 3}{\\lim}\\ln e^{3x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (11-20), use direct substitution to show that each limit leads to the indeterminate form [latex]\\frac{0}{0}[\/latex]. Then, evaluate the limit.<\/p>\r\n\r\n

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11.\u00a0<\/strong>[latex]\\underset{x\\to 4}{\\lim}\\dfrac{x^2-16}{x-4}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572482694\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572482694\"]\r\n

[latex]\\underset{x\\to 4}{\\lim}\\frac{x^2-16}{x-4}=\\frac{16-16}{4-4}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 4}{\\lim}\\frac{x^2-16}{x-4}=\\underset{x\\to 4}{\\lim}\\frac{(x+4)(x-4)}{x-4}=8[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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12.\u00a0<\/strong>[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x-2}{x^2-2x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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13.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}\\dfrac{3x-18}{2x-12}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571586209\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571586209\"][latex]\\underset{x\\to 6}{\\lim}\\frac{3x-18}{2x-12}=\\frac{18-18}{12-12}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 6}{\\lim}\\frac{3x-18}{2x-12}=\\underset{x\\to 6}{\\lim}\\frac{3(x-6)}{2(x-6)}=\\frac{3}{2}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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14.\u00a0<\/strong>[latex]\\underset{h\\to 0}{\\lim}\\dfrac{(1+h)^2-1}{h}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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15.\u00a0<\/strong>[latex]\\underset{t\\to 9}{\\lim}\\dfrac{t-9}{\\sqrt{t}-3}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572499942\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572499942\"]\r\n

[latex]\\underset{t \\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}=\\frac{9-9}{3-3}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{t\\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}=\\underset{t\\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}\\frac{\\sqrt{t}+3}{\\sqrt{t}+3}=\\underset{t\\to 9}{\\lim}(\\sqrt{t}+3)=6[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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16.\u00a0<\/strong>[latex]\\underset{h\\to 0}{\\lim}\\dfrac{\\frac{1}{a+h}-\\frac{1}{a}}{h}[\/latex], where [latex]a[\/latex] is a real-valued constant<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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17.\u00a0<\/strong>[latex]\\underset{\\theta \\to \\pi}{\\lim}\\dfrac{\\sin \\theta}{\\tan \\theta}[\/latex]<\/p>\r\n\r\n

\r\n\r\n[reveal-answer q=\"fs-id1170571657432\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571657432\"][latex]\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\tan \\theta}=\\frac{\\sin \\pi}{\\tan \\pi}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\tan \\theta}=\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\frac{\\sin \\theta}{\\cos \\theta}}=\\underset{\\theta \\to \\pi}{\\lim}\\cos \\theta =-1[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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18.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{x^3-1}{x^2-1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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19.\u00a0<\/strong>[latex]\\underset{x\\to 1\/2}{\\lim}\\dfrac{2x^2+3x-2}{2x-1}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572551526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572551526\"]\r\n

[latex]\\underset{x\\to 1\/2}{\\lim}\\frac{2x^2+3x-2}{2x-1}=\\frac{\\frac{1}{2}+\\frac{3}{2}-2}{1-1}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 1\/2}{\\lim}\\frac{2x^2+3x-2}{2x-1}=\\underset{x\\to 1\/2}{\\lim}\\frac{(2x-1)(x+2)}{2x-1}=\\frac{5}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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20.\u00a0<\/strong>[latex]\\underset{x\\to -3}{\\lim}\\dfrac{\\sqrt{x+4}-1}{x+3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (21-24), use direct substitution to obtain an undefined expression. Then, use the method of (Figure)<\/a> to simplify the function to help determine the limit.<\/p>\r\n\r\n

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21.\u00a0<\/strong>[latex]\\underset{x\\to -2^-}{\\lim}\\dfrac{2x^2+7x-4}{x^2+x-2}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572174724\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572174724\"]\r\n

[latex]-\\infty[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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22.\u00a0<\/strong>[latex]\\underset{x\\to -2^+}{\\lim}\\dfrac{2x^2+7x-4}{x^2+x-2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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23.\u00a0<\/strong>[latex]\\underset{x\\to 1^-}{\\lim}\\dfrac{2x^2+7x-4}{x^2+x-2}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571610806\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571610806\"]\r\n

[latex]-\\infty[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

24. <\/strong>[latex]\\underset{x\\to 1^+}{\\lim}\\dfrac{2x^2+7x-4}{x^2+x-2}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (25-32), assume that [latex]\\underset{x\\to 6}{\\lim}f(x)=4, \\, \\underset{x\\to 6}{\\lim}g(x)=9[\/latex], and [latex]\\underset{x\\to 6}{\\lim}h(x)=6[\/latex]. Use these three facts and the limit laws to evaluate each limit.<\/p>\r\n\r\n

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25.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}2f(x)g(x)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571669784\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571669784\"]\r\n

[latex]\\underset{x\\to 6}{\\lim}2f(x)g(x)=2\\underset{x\\to 6}{\\lim}f(x)\\underset{x\\to 6}{\\lim}g(x)=72[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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26.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}\\dfrac{g(x)-1}{f(x)}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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27.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}(f(x)+\\frac{1}{3}g(x))[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572480532\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572480532\"]\r\n

[latex]\\underset{x\\to 6}{\\lim}(f(x)+\\frac{1}{3}g(x))=\\underset{x\\to 6}{\\lim}f(x)+\\frac{1}{3}\\underset{x\\to 6}{\\lim}g(x)=7[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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28.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}\\dfrac{(h(x))^3}{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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29.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}\\sqrt{g(x)-f(x)}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572217368\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572217368\"]\r\n

[latex]\\underset{x\\to 6}{\\lim}\\sqrt{g(x)-f(x)}=\\sqrt{\\underset{x\\to 6}{\\lim}g(x)-\\underset{x\\to 6}{\\lim}f(x)}=\\sqrt{5}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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30.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}x \\cdot h(x)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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31.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}[(x+1)\\cdot f(x)][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572549082\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572549082\"][latex]\\underset{x\\to 6}{\\lim}[(x+1)\\cdot f(x)]=(\\underset{x\\to 6}{\\lim}(x+1))(\\underset{x\\to 6}{\\lim}f(x))=28[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

32.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}(f(x)\\cdot g(x)-h(x))[\/latex]<\/div>\r\nIn the following exercises (33-35), use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.\r\n
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\r\n\r\n33. [T]\u00a0<\/strong>[latex]f(x)=\\begin{cases} x^2, & x \\le 3 \\\\ x+4, & x > 3 \\end{cases}[\/latex]\r\n
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  1. [latex]\\underset{x\\to 3^-}{\\lim}f(x)[\/latex]<\/li>\r\n \t
  2. [latex]\\underset{x\\to 3^+}{\\lim}f(x)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572624250\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572624250\"]\"The<\/span>\r\na. 9; b. 7[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n
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    34. [T]\u00a0<\/strong>[latex]g(x)=\\begin{cases} x^3 - 1, & x \\le 0 \\\\ 1, & x > 0 \\end{cases}[\/latex]<\/p>\r\n\r\n

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    1. [latex]\\underset{x\\to 0^-}{\\lim}g(x)[\/latex]<\/li>\r\n \t
    2. [latex]\\underset{x\\to 0^+}{\\lim}g(x)[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
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      35. [T]\u00a0<\/strong>[latex]h(x)=\\begin{cases} x^2-2x+1, & x < 2 \\\\ 3 - x, & x \\ge 2 \\end{cases}[\/latex]<\/p>\r\n\r\n

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        1. [latex]\\underset{x\\to 2^-}{\\lim}h(x)[\/latex]<\/li>\r\n \t
        2. [latex]\\underset{x\\to 2^+}{\\lim}h(x)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170572268044\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572268044\"]\"The<\/span>\r\na. 1; b. 1[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

          In the following exercises (36-43), use the graphs below and the limit laws to evaluate each limit.<\/p>\r\n\"Two<\/span>\r\n

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          36.\u00a0<\/strong>[latex]\\underset{x\\to -3^+}{\\lim}(f(x)+g(x))[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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          37.\u00a0<\/strong>[latex]\\underset{x\\to -3^-}{\\lim}(f(x)-3g(x))[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572434876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572434876\"]\r\n

          [latex]\\underset{x\\to -3^-}{\\lim}(f(x)-3g(x))=\\underset{x\\to -3^-}{\\lim}f(x)-3\\underset{x\\to -3^-}{\\lim}g(x)=0+6=6[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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          38.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{f(x)g(x)}{3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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          39.\u00a0<\/strong>[latex]\\underset{x\\to -5}{\\lim}\\dfrac{2+g(x)}{f(x)}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572219572\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572219572\"]\r\n

          [latex]\\underset{x\\to -5}{\\lim}\\frac{2+g(x)}{f(x)}=\\frac{2+(\\underset{x\\to -5}{\\lim}g(x))}{\\underset{x\\to -5}{\\lim}f(x)}=\\frac{2+0}{2}=1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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          40.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}(f(x))^2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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          41.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}\\sqrt[3]{f(x)-g(x)}[\/latex]<\/p>\r\n[reveal-answer q=\"957757\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"957757\"]\r\n[latex]\\underset{x\\to 1}{\\lim}\\sqrt[3]{f(x)-g(x)}=\\sqrt[3]{\\underset{x\\to 1}{\\lim}f(x)-\\underset{x\\to 1}{\\lim}g(x)}=\\sqrt[3]{2+5}=\\sqrt[3]{7}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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          42.\u00a0<\/strong>[latex]\\underset{x\\to -7}{\\lim}(x \\cdot g(x))[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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          \r\n\r\n43.\u00a0<\/strong>[latex]\\underset{x\\to -9}{\\lim}[xf(x)+2g(x)][\/latex]\r\n\r\n[reveal-answer q=\"fs-id1170572540774\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572540774\"]\r\n

          [latex]\\underset{x\\to -9}{\\lim}(x\\cdot f(x)+2g(x))=(\\underset{x\\to -9}{\\lim}x)(\\underset{x\\to -9}{\\lim}f(x))+2\\underset{x\\to -9}{\\lim}(g(x))=(-9)(6)+2(4)=-46[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

          For the following problems (44-46), evaluate the limit using the Squeeze Theorem. Use a calculator to graph the functions [latex]f(x), \\, g(x)[\/latex], and [latex]h(x)[\/latex] when possible.<\/p>\r\n\r\n

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          44. [T]<\/strong> True or False? If [latex]2x-1\\le g(x)\\le x^2-2x+3[\/latex], then [latex]\\underset{x\\to 2}{\\lim}g(x)=0[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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          45. [T]\u00a0<\/strong>[latex]\\underset{\\theta \\to 0}{\\lim}\\theta^2 \\cos\\left(\\frac{1}{\\theta}\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571625945\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571625945\"]\r\n

          The limit is zero.<\/p>\r\n\"The<\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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          46.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\lim}f(x)[\/latex], where [latex]f(x)=\\begin{cases} 0, & x \\, \\text{rational} \\\\ x^2, & x \\, \\text{irrational} \\end{cases}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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          47. [T]<\/strong> In physics, the magnitude of an electric field generated by a point charge at a distance [latex]r[\/latex] in vacuum is governed by Coulomb\u2019s law: [latex]E(r)=\\large \\frac{q}{4\\pi \\varepsilon_0 r^2}[\/latex], where [latex]E[\/latex] represents the magnitude of the electric field, [latex]q[\/latex] is the charge of the particle, [latex]r[\/latex] is the distance between the particle and where the strength of the field is measured, and [latex]\\large \\frac{1}{4\\pi \\varepsilon_0}[\/latex] is Coulomb\u2019s constant: [latex]8.988 \\times 10^9 \\, \\text{N} \\cdot \\text{m}^2\/\\text{C}^2[\/latex].<\/p>\r\n\r\n

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          1. Use a graphing calculator to graph [latex]E(r)[\/latex] given that the charge of the particle is [latex]q=10^{-10}[\/latex].<\/li>\r\n \t
          2. Evaluate [latex]\\underset{r\\to 0^+}{\\lim}E(r)[\/latex]. What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170571612256\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571612256\"]\r\n

            a.<\/p>\r\n\"A<\/span>\r\nb. [latex]\\underset{r\\to 0^+}{\\lim}E(r)=\\infty[\/latex]. The magnitude of the electric field as you approach the particle [latex]q[\/latex] becomes infinite. It does not make physical sense to evaluate negative distance.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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            48. [T]<\/strong> The density of an object is given by its mass divided by its volume: [latex]\\rho =\\dfrac{m}{V}[\/latex].<\/p>\r\n\r\n

              \r\n \t
            1. Use a calculator to plot the volume as a function of density [latex](V=\\frac{m}{\\rho})[\/latex], assuming you are examining something of mass 8 kg ([latex]m=8[\/latex]).<\/li>\r\n \t
            2. Evaluate [latex]\\underset{\\rho \\to 0^+}{\\lim}V(\\rho)[\/latex] and explain the physical meaning.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

              In the following exercises (1-4), use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).<\/p>\n

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              1.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\lim}(4x^2-2x+3)[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              Use constant multiple law and difference law: [latex]\\underset{x\\to 0}{\\lim}(4x^2-2x+3)=4\\underset{x\\to 0}{\\lim}x^2-2\\underset{x\\to 0}{\\lim}x+\\underset{x\\to 0}{\\lim}3=3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              2.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}\\frac{x^3+3x^2+5}{4-7x}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              3.\u00a0<\/strong>[latex]\\underset{x\\to -2}{\\lim}\\sqrt{x^2-6x+3}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              Use root law: [latex]\\underset{x\\to -2}{\\lim}\\sqrt{x^2-6x+3}=\\sqrt{\\underset{x\\to -2}{\\lim}(x^2-6x+3)}=\\sqrt{19}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              4.\u00a0<\/strong>[latex]\\underset{x\\to -1}{\\lim}(9x+1)^2[\/latex]<\/p>\n<\/div>\n<\/div>\n

              In the following exercises (5-10), use direct substitution to evaluate each limit.<\/p>\n

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              5.\u00a0<\/strong>[latex]\\underset{x\\to 7}{\\lim}x^2[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              49<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              6.\u00a0<\/strong>[latex]\\underset{x\\to -2}{\\lim}(4x^2-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              7.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{1}{1+ \\sin x}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              1<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              8.\u00a0<\/strong>[latex]\\underset{x\\to 2}{\\lim}e^{2x-x^2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              9.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{2-7x}{x+6}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
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              [latex]-\\frac{5}{7}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              10.\u00a0<\/strong>[latex]\\underset{x\\to 3}{\\lim}\\ln e^{3x}[\/latex]<\/p>\n<\/div>\n<\/div>\n

              In the following exercises (11-20), use direct substitution to show that each limit leads to the indeterminate form [latex]\\frac{0}{0}[\/latex]. Then, evaluate the limit.<\/p>\n

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              11.\u00a0<\/strong>[latex]\\underset{x\\to 4}{\\lim}\\dfrac{x^2-16}{x-4}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              [latex]\\underset{x\\to 4}{\\lim}\\frac{x^2-16}{x-4}=\\frac{16-16}{4-4}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 4}{\\lim}\\frac{x^2-16}{x-4}=\\underset{x\\to 4}{\\lim}\\frac{(x+4)(x-4)}{x-4}=8[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              12.\u00a0<\/strong>[latex]\\underset{x\\to 2}{\\lim}\\dfrac{x-2}{x^2-2x}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              13.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}\\dfrac{3x-18}{2x-12}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              [latex]\\underset{x\\to 6}{\\lim}\\frac{3x-18}{2x-12}=\\frac{18-18}{12-12}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 6}{\\lim}\\frac{3x-18}{2x-12}=\\underset{x\\to 6}{\\lim}\\frac{3(x-6)}{2(x-6)}=\\frac{3}{2}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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              14.\u00a0<\/strong>[latex]\\underset{h\\to 0}{\\lim}\\dfrac{(1+h)^2-1}{h}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              15.\u00a0<\/strong>[latex]\\underset{t\\to 9}{\\lim}\\dfrac{t-9}{\\sqrt{t}-3}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              [latex]\\underset{t \\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}=\\frac{9-9}{3-3}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{t\\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}=\\underset{t\\to 9}{\\lim}\\frac{t-9}{\\sqrt{t}-3}\\frac{\\sqrt{t}+3}{\\sqrt{t}+3}=\\underset{t\\to 9}{\\lim}(\\sqrt{t}+3)=6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              16.\u00a0<\/strong>[latex]\\underset{h\\to 0}{\\lim}\\dfrac{\\frac{1}{a+h}-\\frac{1}{a}}{h}[\/latex], where [latex]a[\/latex] is a real-valued constant<\/p>\n<\/div>\n<\/div>\n

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              17.\u00a0<\/strong>[latex]\\underset{\\theta \\to \\pi}{\\lim}\\dfrac{\\sin \\theta}{\\tan \\theta}[\/latex]<\/p>\n

              \n
              Show Solution<\/span><\/p>\n
              [latex]\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\tan \\theta}=\\frac{\\sin \\pi}{\\tan \\pi}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\tan \\theta}=\\underset{\\theta \\to \\pi}{\\lim}\\frac{\\sin \\theta}{\\frac{\\sin \\theta}{\\cos \\theta}}=\\underset{\\theta \\to \\pi}{\\lim}\\cos \\theta =-1[\/latex].\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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              18.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}\\dfrac{x^3-1}{x^2-1}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              19.\u00a0<\/strong>[latex]\\underset{x\\to 1\/2}{\\lim}\\dfrac{2x^2+3x-2}{2x-1}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              [latex]\\underset{x\\to 1\/2}{\\lim}\\frac{2x^2+3x-2}{2x-1}=\\frac{\\frac{1}{2}+\\frac{3}{2}-2}{1-1}=\\frac{0}{0}[\/latex]; then, [latex]\\underset{x\\to 1\/2}{\\lim}\\frac{2x^2+3x-2}{2x-1}=\\underset{x\\to 1\/2}{\\lim}\\frac{(2x-1)(x+2)}{2x-1}=\\frac{5}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              20.\u00a0<\/strong>[latex]\\underset{x\\to -3}{\\lim}\\dfrac{\\sqrt{x+4}-1}{x+3}[\/latex]<\/p>\n<\/div>\n<\/div>\n

              In the following exercises (21-24), use direct substitution to obtain an undefined expression. Then, use the method of (Figure)<\/a> to simplify the function to help determine the limit.<\/p>\n

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              21.\u00a0<\/strong>[latex]\\underset{x\\to -2^-}{\\lim}\\dfrac{2x^2+7x-4}{x^2+x-2}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              [latex]-\\infty[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              22.\u00a0<\/strong>[latex]\\underset{x\\to -2^+}{\\lim}\\dfrac{2x^2+7x-4}{x^2+x-2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              23.\u00a0<\/strong>[latex]\\underset{x\\to 1^-}{\\lim}\\dfrac{2x^2+7x-4}{x^2+x-2}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              [latex]-\\infty[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

              24. <\/strong>[latex]\\underset{x\\to 1^+}{\\lim}\\dfrac{2x^2+7x-4}{x^2+x-2}[\/latex]<\/div>\n<\/div>\n<\/div>\n

              In the following exercises (25-32), assume that [latex]\\underset{x\\to 6}{\\lim}f(x)=4, \\, \\underset{x\\to 6}{\\lim}g(x)=9[\/latex], and [latex]\\underset{x\\to 6}{\\lim}h(x)=6[\/latex]. Use these three facts and the limit laws to evaluate each limit.<\/p>\n

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              25.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}2f(x)g(x)[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              [latex]\\underset{x\\to 6}{\\lim}2f(x)g(x)=2\\underset{x\\to 6}{\\lim}f(x)\\underset{x\\to 6}{\\lim}g(x)=72[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              26.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}\\dfrac{g(x)-1}{f(x)}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              27.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}(f(x)+\\frac{1}{3}g(x))[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              [latex]\\underset{x\\to 6}{\\lim}(f(x)+\\frac{1}{3}g(x))=\\underset{x\\to 6}{\\lim}f(x)+\\frac{1}{3}\\underset{x\\to 6}{\\lim}g(x)=7[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              28.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}\\dfrac{(h(x))^3}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              29.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}\\sqrt{g(x)-f(x)}[\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              \n

              [latex]\\underset{x\\to 6}{\\lim}\\sqrt{g(x)-f(x)}=\\sqrt{\\underset{x\\to 6}{\\lim}g(x)-\\underset{x\\to 6}{\\lim}f(x)}=\\sqrt{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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              30.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}x \\cdot h(x)[\/latex]<\/p>\n<\/div>\n<\/div>\n

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              31.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}[(x+1)\\cdot f(x)][\/latex]<\/p>\n

              Show Solution<\/span><\/p>\n
              [latex]\\underset{x\\to 6}{\\lim}[(x+1)\\cdot f(x)]=(\\underset{x\\to 6}{\\lim}(x+1))(\\underset{x\\to 6}{\\lim}f(x))=28[\/latex].\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
              32.\u00a0<\/strong>[latex]\\underset{x\\to 6}{\\lim}(f(x)\\cdot g(x)-h(x))[\/latex]<\/div>\n

              In the following exercises (33-35), use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.<\/p>\n

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              33. [T]\u00a0<\/strong>[latex]f(x)=\\begin{cases} x^2, & x \\le 3 \\\\ x+4, & x > 3 \\end{cases}[\/latex]<\/p>\n

                \n
              1. [latex]\\underset{x\\to 3^-}{\\lim}f(x)[\/latex]<\/li>\n
              2. [latex]\\underset{x\\to 3^+}{\\lim}f(x)[\/latex]<\/li>\n<\/ol>\n
                Show Solution<\/span><\/p>\n
                \"The<\/span>
                \na. 9; b. 7<\/div>\n<\/div>\n<\/div>\n<\/div>\n
                \n
                \n

                34. [T]\u00a0<\/strong>[latex]g(x)=\\begin{cases} x^3 - 1, & x \\le 0 \\\\ 1, & x > 0 \\end{cases}[\/latex]<\/p>\n

                  \n
                1. [latex]\\underset{x\\to 0^-}{\\lim}g(x)[\/latex]<\/li>\n
                2. [latex]\\underset{x\\to 0^+}{\\lim}g(x)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
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                  35. [T]\u00a0<\/strong>[latex]h(x)=\\begin{cases} x^2-2x+1, & x < 2 \\\\ 3 - x, & x \\ge 2 \\end{cases}[\/latex]<\/p>\n

                    \n
                  1. \n
                      \n
                    1. [latex]\\underset{x\\to 2^-}{\\lim}h(x)[\/latex]<\/li>\n
                    2. [latex]\\underset{x\\to 2^+}{\\lim}h(x)[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                      Show Solution<\/span><\/p>\n
                      \"The<\/span>
                      \na. 1; b. 1<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                      In the following exercises (36-43), use the graphs below and the limit laws to evaluate each limit.<\/p>\n

                      \"Two<\/span><\/p>\n

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                      36.\u00a0<\/strong>[latex]\\underset{x\\to -3^+}{\\lim}(f(x)+g(x))[\/latex]<\/p>\n<\/div>\n<\/div>\n

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                      37.\u00a0<\/strong>[latex]\\underset{x\\to -3^-}{\\lim}(f(x)-3g(x))[\/latex]<\/p>\n

                      Show Solution<\/span><\/p>\n
                      \n

                      [latex]\\underset{x\\to -3^-}{\\lim}(f(x)-3g(x))=\\underset{x\\to -3^-}{\\lim}f(x)-3\\underset{x\\to -3^-}{\\lim}g(x)=0+6=6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                      38.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\lim}\\dfrac{f(x)g(x)}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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                      39.\u00a0<\/strong>[latex]\\underset{x\\to -5}{\\lim}\\dfrac{2+g(x)}{f(x)}[\/latex]<\/p>\n

                      Show Solution<\/span><\/p>\n
                      \n

                      [latex]\\underset{x\\to -5}{\\lim}\\frac{2+g(x)}{f(x)}=\\frac{2+(\\underset{x\\to -5}{\\lim}g(x))}{\\underset{x\\to -5}{\\lim}f(x)}=\\frac{2+0}{2}=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                      40.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}(f(x))^2[\/latex]<\/p>\n<\/div>\n<\/div>\n

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                      41.\u00a0<\/strong>[latex]\\underset{x\\to 1}{\\lim}\\sqrt[3]{f(x)-g(x)}[\/latex]<\/p>\n

                      Show Solution<\/span><\/p>\n
                      \n[latex]\\underset{x\\to 1}{\\lim}\\sqrt[3]{f(x)-g(x)}=\\sqrt[3]{\\underset{x\\to 1}{\\lim}f(x)-\\underset{x\\to 1}{\\lim}g(x)}=\\sqrt[3]{2+5}=\\sqrt[3]{7}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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                      42.\u00a0<\/strong>[latex]\\underset{x\\to -7}{\\lim}(x \\cdot g(x))[\/latex]<\/p>\n<\/div>\n<\/div>\n

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                      43.\u00a0<\/strong>[latex]\\underset{x\\to -9}{\\lim}[xf(x)+2g(x)][\/latex]<\/p>\n

                      Show Solution<\/span><\/p>\n
                      \n

                      [latex]\\underset{x\\to -9}{\\lim}(x\\cdot f(x)+2g(x))=(\\underset{x\\to -9}{\\lim}x)(\\underset{x\\to -9}{\\lim}f(x))+2\\underset{x\\to -9}{\\lim}(g(x))=(-9)(6)+2(4)=-46[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                      For the following problems (44-46), evaluate the limit using the Squeeze Theorem. Use a calculator to graph the functions [latex]f(x), \\, g(x)[\/latex], and [latex]h(x)[\/latex] when possible.<\/p>\n

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                      44. [T]<\/strong> True or False? If [latex]2x-1\\le g(x)\\le x^2-2x+3[\/latex], then [latex]\\underset{x\\to 2}{\\lim}g(x)=0[\/latex].<\/p>\n<\/div>\n<\/div>\n

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                      45. [T]\u00a0<\/strong>[latex]\\underset{\\theta \\to 0}{\\lim}\\theta^2 \\cos\\left(\\frac{1}{\\theta}\\right)[\/latex]<\/p>\n

                      Show Solution<\/span><\/p>\n
                      \n

                      The limit is zero.<\/p>\n

                      \"The<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                      46.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\lim}f(x)[\/latex], where [latex]f(x)=\\begin{cases} 0, & x \\, \\text{rational} \\\\ x^2, & x \\, \\text{irrational} \\end{cases}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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                      47. [T]<\/strong> In physics, the magnitude of an electric field generated by a point charge at a distance [latex]r[\/latex] in vacuum is governed by Coulomb\u2019s law: [latex]E(r)=\\large \\frac{q}{4\\pi \\varepsilon_0 r^2}[\/latex], where [latex]E[\/latex] represents the magnitude of the electric field, [latex]q[\/latex] is the charge of the particle, [latex]r[\/latex] is the distance between the particle and where the strength of the field is measured, and [latex]\\large \\frac{1}{4\\pi \\varepsilon_0}[\/latex] is Coulomb\u2019s constant: [latex]8.988 \\times 10^9 \\, \\text{N} \\cdot \\text{m}^2\/\\text{C}^2[\/latex].<\/p>\n

                        \n
                      1. Use a graphing calculator to graph [latex]E(r)[\/latex] given that the charge of the particle is [latex]q=10^{-10}[\/latex].<\/li>\n
                      2. Evaluate [latex]\\underset{r\\to 0^+}{\\lim}E(r)[\/latex]. What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?<\/li>\n<\/ol>\n
                        Show Solution<\/span><\/p>\n
                        \n

                        a.<\/p>\n

                        \"A<\/span>
                        \nb. [latex]\\underset{r\\to 0^+}{\\lim}E(r)=\\infty[\/latex]. The magnitude of the electric field as you approach the particle [latex]q[\/latex] becomes infinite. It does not make physical sense to evaluate negative distance.<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                        48. [T]<\/strong> The density of an object is given by its mass divided by its volume: [latex]\\rho =\\dfrac{m}{V}[\/latex].<\/p>\n

                          \n
                        1. Use a calculator to plot the volume as a function of density [latex](V=\\frac{m}{\\rho})[\/latex], assuming you are examining something of mass 8 kg ([latex]m=8[\/latex]).<\/li>\n
                        2. Evaluate [latex]\\underset{\\rho \\to 0^+}{\\lim}V(\\rho)[\/latex] and explain the physical meaning.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t
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