{"id":3139,"date":"2024-06-13T16:32:03","date_gmt":"2024-06-13T16:32:03","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&p=3139"},"modified":"2024-08-05T13:31:22","modified_gmt":"2024-08-05T13:31:22","slug":"lhopitals-rule-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/lhopitals-rule-learn-it-2\/","title":{"raw":"L\u2019H\u00f4pital\u2019s Rule: Learn It 2","rendered":"L\u2019H\u00f4pital\u2019s Rule: Learn It 2"},"content":{"raw":"

L\u2019H\u00f4pital\u2019s Rule<\/h2>\r\n

Applying L\u2019H\u00f4pital\u2019s Rule Cont.<\/h3>\r\n

L\u2019H\u00f4pital\u2019s Rule ([latex]\\infty \/ \\infty[\/latex] Case)<\/h4>\r\n

L'H\u00f4pital's Rule provides a method to resolve indeterminate forms in calculus, specifically those that result in [latex]\\infty \/ \\infty[\/latex] when calculating limits.<\/p>\r\n

\r\n

L\u2019H\u00f4pital\u2019s rule ([latex]\\infty \/ \\infty[\/latex] case)<\/h3>\r\n

Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval containing [latex]a[\/latex], except possibly at [latex]a[\/latex]. Suppose [latex]\\underset{x\\to a}{\\lim}f(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]) and [latex]\\underset{x\\to a}{\\lim}g(x)=\\infty[\/latex] (or [latex]\u2212\\infty[\/latex]). Then,<\/p>\r\n

[latex]\\underset{x\\to a}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to a}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex],<\/div>\r\n

assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex]. This result also holds if the limit is infinite, if [latex]a=\\infty[\/latex] or [latex]\u2212\\infty[\/latex], or the limit is one-sided.<\/p>\r\n<\/section>\r\n

\r\n

Evaluate each of the following limits by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n

    \r\n\t
  1. [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}[\/latex]<\/li>\r\n\t
  2. [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}[\/latex]<\/li>\r\n<\/ol>\r\n

    [reveal-answer q=\"fs-id1165042376758\"]Show Solution[\/reveal-answer]
    \r\n[hidden-answer a=\"fs-id1165042376758\"]<\/p>\r\n

      \r\n\t
    1. Since [latex]3x+5[\/latex] and [latex]2x+1[\/latex] are first-degree polynomials with positive leading coefficients, [latex]\\underset{x\\to \\infty }{\\lim}(3x+5)=\\infty [\/latex] and [latex]\\underset{x\\to \\infty }{\\lim}(2x+1)=\\infty[\/latex]. Therefore, we apply L\u2019H\u00f4pital\u2019s rule and obtain
      \r\n
      [latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{3x+5}{2x+\\frac{1}{x}}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3}{2}=\\dfrac{3}{2}[\/latex].<\/div>\r\n

      Note that this limit can also be calculated without invoking L\u2019H\u00f4pital\u2019s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[\/latex] in the denominator. In doing so, we saw that<\/p>\r\n

      [latex]\\underset{x\\to \\infty }{\\lim}\\dfrac{3x+5}{2x+1}=\\underset{x\\to \\infty }{\\lim}\\dfrac{3+\\frac{5}{x}}{2+\\frac{1}{x}}=\\dfrac{3}{2}[\/latex].<\/div>\r\n

      L\u2019H\u00f4pital\u2019s rule provides us with an alternative means of evaluating this type of limit.<\/p>\r\n<\/li>\r\n\t

    2. Here, [latex]\\underset{x\\to 0^+}{\\lim} \\ln x=\u2212\\infty [\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} \\cot x=\\infty[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule and obtain
      \r\n
      [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\frac{1}{x}}{\u2212\\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}[\/latex].<\/div>\r\n

      Now as [latex]x\\to 0^+[\/latex], [latex]\\csc^2 x\\to \\infty[\/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\\csc x[\/latex] to write<\/p>\r\n

      [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{1}{\u2212x \\csc^2 x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}[\/latex].<\/div>\r\n

      Now [latex]\\underset{x\\to 0^+}{\\lim} \\sin^2 x=0[\/latex] and [latex]\\underset{x\\to 0^+}{\\lim} x=0[\/latex], so we apply L\u2019H\u00f4pital\u2019s rule again. We find<\/p>\r\n

      [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\sin^2 x}{\u2212x}=\\underset{x\\to 0^+}{\\lim}\\dfrac{2 \\sin x \\cos x}{-1}=\\dfrac{0}{-1}=0[\/latex].<\/div>\r\n

      We conclude that<\/p>\r\n

      [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\ln x}{\\cot x}=0[\/latex].<\/div>\r\n<\/li>\r\n<\/ol>\r\n

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

      \r\n

      To correctly apply L'H\u00f4pital's Rule to a quotient [latex]\\frac{f(x)}{g(x)}[\/latex], it is essential that the original limit of the quotient is an indeterminate form, either [latex]0\/0[\/latex] or [latex]\\infty \/ \\infty[\/latex]. This is crucial because applying the rule outside these conditions does not yield valid results.<\/p>\r\n<\/section>\r\n

      While L'H\u00f4pital's Rule is an invaluable tool for calculus, its application should be carefully considered. Not all limits of the form [latex]\\infty \/ \\infty[\/latex] are suitable for L'H\u00f4pital's Rule without additional analysis or transformation of the function.<\/p>\r\n

      Consider the following non-applicable example to better understand the limitations:<\/p>\r\n

      \r\n

      Consider [latex]\\underset{x\\to 1}{\\lim}\\dfrac{x^2+5}{3x+4}[\/latex]. Show that the limit cannot be evaluated by applying L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n

      Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L\u2019H\u00f4pital\u2019s rule. If we try to do so, we get<\/p>\r\n

      [latex]\\frac{d}{dx}(x^2+5)=2x[\/latex]<\/div>\r\n

      and,<\/p>\r\n

      [latex]\\frac{d}{dx}(3x+4)=3[\/latex]<\/div>\r\n

      At which point we would conclude erroneously that<\/p>\r\n

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

      However, since [latex]\\underset{x\\to 1}{\\lim}(x^2+5)=6[\/latex] and [latex]\\underset{x\\to 1}{\\lim}(3x+4)=7[\/latex], we actually have:<\/p>\r\n

      [latex]\\underset{x\\to 1}{\\lim}\\frac{x^2+5}{3x+4}=\\frac{6}{7}[\/latex]<\/div>\r\n

      We can conclude that<\/p>\r\n

      [latex]\\underset{x\\to 1}{\\lim}\\frac{x^2+5}{3x+4}\\ne \\underset{x\\to 1}{\\lim}\\frac{\\frac{d}{dx}(x^2+5)}{\\frac{d}{dx}(3x+4)}[\/latex].[\/hidden-answer]<\/div>\r\n<\/section>\r\n
      \r\n

      Explain why we cannot apply L\u2019H\u00f4pital\u2019s rule to evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\cos x}{x}[\/latex]. Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\dfrac{\\cos x}{x}[\/latex] by other means.<\/p>\r\n

      [reveal-answer q=\"6688909\"]Hint[\/reveal-answer]
      \r\n[hidden-answer a=\"6688909\"]<\/p>\r\n

      Determine the limits of the numerator and denominator separately.<\/p>\r\n

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

      [reveal-answer q=\"fs-id1165042632518\"]Show Solution[\/reveal-answer]
      \r\n[hidden-answer a=\"fs-id1165042632518\"]<\/p>\r\n

      [latex]\\underset{x\\to 0^+}{\\lim} \\cos x=1[\/latex]. Therefore, we cannot apply L\u2019H\u00f4pital\u2019s rule. The limit of the quotient is [latex]\\infty [\/latex]<\/p>\r\n

      Watch the following video to see the worked solution to this example.<\/p>\r\n