L’Hôpital’s Rule

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Evaluate each of the following limits by applying L’Hôpital’s rule.

   1. \(\underset{x\to 0}{\text{lim}}\frac{1-\text{cos}\ x}{x}\)
   2. \(\underset{x\to 1}{\text{lim}}\frac{\text{sin}(\pi x)}{\text{ln}\ x}\)
   3. \(\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{1\text{/}x}\)
   4. \(\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x-x}{{x}^{2}}\)
   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   1. Since the numerator \(1-\text{cos}\ x\to 0\) and the denominator \(x\to 0,\) we can apply L’Hôpital’s rule to evaluate this limit. We have
      \[\begin{array}{ll}\underset{x\to 0}{\text{lim}}\frac{1-\text{cos}\ x}{x} & =\underset{x\to 0}{\text{lim}}\frac{\frac{d}{dx}(1-\text{cos}\ x)}{\frac{d}{dx}(x)} \\ & =\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x}{1} \\ & =\frac{\underset{x\to 0}{\text{lim}}(\text{sin}\ x)}{\underset{x\to 0}{\text{lim}}(1)} \\ & =\frac{0}{1}=0.\end{array}\]
   2. As \(x\to 1,\) the numerator \(\text{sin}(\pi x)\to 0\) and the denominator \(\text{ln}(x)\to 0.\) Therefore, we can apply L’Hôpital’s rule. We obtain
      \[\begin{array}{ll}\underset{x\to 1}{\text{lim}}\frac{\text{sin}(\pi x)}{\text{ln}\ x} & =\underset{x\to 1}{\text{lim}}\frac{\pi \ \text{cos}(\pi x)}{1\text{/}x} \\ & =\underset{x\to 1}{\text{lim}}(\pi x)\text{cos}(\pi x) \\ & =(\pi \cdot 1)(-1)=\text{-}\pi .\end{array}\]
   3. As \(x\to \infty ,\) the numerator \({e}^{1\text{/}x}-1\to 0\) and the denominator \((\frac{1}{x})\to 0.\) Therefore, we can apply L’Hôpital’s rule. We obtain
      \[\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}-1}{\frac{1}{x}}=\underset{x\to \infty }{\text{lim}}\frac{{e}^{1\text{/}x}(\frac{-1}{{x}^{2}})}{(\frac{-1}{{x}^{2}})}=\underset{x\to \infty }{\text{lim}}{e}^{1\text{/}x}={e}^{0}=1.\]
   4. As \(x\to 0,\) both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
      \[\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x-x}{{x}^{2}}=\underset{x\to 0}{\text{lim}}\frac{\text{cos}\ x-1}{2x}.\]
      Since the numerator and denominator of this new quotient both approach zero as \(x\to 0,\) we apply L’Hôpital’s rule again. In doing so, we see that
      \[\underset{x\to 0}{\text{lim}}\frac{\text{cos}\ x-1}{2x}=\underset{x\to 0}{\text{lim}}\frac{\text{-}\text{sin}\ x}{2}=0.\]
      Therefore, we conclude that
      \[\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x-x}{{x}^{2}}=0.\]

2. Evaluate \(\underset{x\to 0}{\text{lim}}\frac{x}{\text{tan}\ x}.\)

   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   \(1\)

3. Evaluate each of the following limits by applying L’Hôpital’s rule.

   1. \(\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}\)
   2. \(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{\text{cot}\ x}\)
   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   1. Since \(3x+5\) and \(2x+1\) are first-degree polynomials with positive leading coefficients, \(\underset{x\to \infty }{\text{lim}}(3x+5)=\infty\) and \(\underset{x\to \infty }{\text{lim}}(2x+1)=\infty .\) Therefore, we apply L’Hôpital’s rule and obtain
      \[\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3}{2}=\frac{3}{2}.\]
      Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of \(x\) in the denominator. In doing so, we saw that
      \[\underset{x\to \infty }{\text{lim}}\frac{3x+5}{2x+1}=\underset{x\to \infty }{\text{lim}}\frac{3+5\text{/}x}{2+1\text{/}x}=\frac{3}{2}.\]
      L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
   2. Here, \(\underset{x\to {0}^{+}}{\text{lim}}\text{ln}\ x=\text{-}\infty\) and \(\underset{x\to {0}^{+}}{\text{lim}}\text{cot}\ x=\infty .\) Therefore, we can apply L’Hôpital’s rule and obtain
      \[\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{\text{cot}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{\text{-}{\text{csc}}^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{-}x\ {\text{csc}}^{2}x}.\]
      Now as \(x\to {0}^{+},\) \({\text{csc}}^{2}x\to \infty .\) 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 \(\text{csc}\ x\) to write
      \[\underset{x\to {0}^{+}}{\text{lim}}\frac{1}{\text{-}x\ {\text{csc}}^{2}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{{\text{sin}}^{2}x}{\text{-}x}.\]
      Now \(\underset{x\to {0}^{+}}{\text{lim}}{\text{sin}}^{2}x=0\) and \(\underset{x\to {0}^{+}}{\text{lim}}x=0,\) so we apply L’Hôpital’s rule again. We find
      \[\underset{x\to {0}^{+}}{\text{lim}}\frac{{\text{sin}}^{2}x}{\text{-}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{2\ \text{sin}\ x\ \text{cos}\ x}{-1}=\frac{0}{-1}=0.\]
      We conclude that
      \[\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{\text{cot}\ x}=0.\]

4. Evaluate \(\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{5x}.\)

   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   \(0\)

5. Consider \(\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}.\) Show that the limit cannot be evaluated by applying L’Hôpital’s rule.

   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get

   \[\frac{d}{dx}({x}^{2}+5)=2x\]

   and

   \[\frac{d}{dx}(3x+4)=3.\]

   At which point we would conclude erroneously that

   \[\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}=\underset{x\to 1}{\text{lim}}\frac{2x}{3}=\frac{2}{3}.\]

   However, since \(\underset{x\to 1}{\text{lim}}({x}^{2}+5)=6\) and \(\underset{x\to 1}{\text{lim}}(3x+4)=7,\) we actually have

   \[\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}=\frac{6}{7}.\]

   We can conclude that

   \[\underset{x\to 1}{\text{lim}}\frac{{x}^{2}+5}{3x+4}\ne \underset{x\to 1}{\text{lim}}\frac{\frac{d}{dx}({x}^{2}+5)}{\frac{d}{dx}(3x+4)}.\]

6. Explain why we cannot apply L’Hôpital’s rule to evaluate \(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{cos}\ x}{x}.\) Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{cos}\ x}{x}\) by other means.

   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   \(\underset{x\to {0}^{+}}{\text{lim}}\text{cos}\ x=1.\) Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is \(\infty\)

7. Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}x\ \text{ln}\ x.\)

   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   First, rewrite the function \(x\ \text{ln}\ x\) as a quotient to apply L’Hôpital’s rule. If we write

   \[x\ \text{ln}\ x=\frac{\text{ln}\ x}{1\text{/}x},\]

   we see that \(\text{ln}\ x\to \text{-}\infty\) as \(x\to {0}^{+}\) and \(\frac{1}{x}\to \infty\) as \(x\to {0}^{+}.\) Therefore, we can apply L’Hôpital’s rule and obtain

   \[\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{1\text{/}x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{\frac{d}{dx}(\text{ln}\ x)}{\frac{d}{dx}(1\text{/}x)}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{-1\text{/}{x}^{2}}=\underset{x\to {0}^{+}}{\text{lim}}(\text{-}x)=0.\]

   We conclude that

   \[\underset{x\to {0}^{+}}{\text{lim}}x\ \text{ln}\ x=0.\]

8. Evaluate \(\underset{x\to 0}{\text{lim}}x\ \text{cot}\ x.\)

   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   \(1\)

9. Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}(\frac{1}{{x}^{2}}-\frac{1}{\text{tan}\ x}).\)

   ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

   By combining the fractions, we can write the function as a quotient. Since the least common denominator is \({x}^{2}\text{tan}\ x,\) we have

   \[\frac{1}{{x}^{2}}-\frac{1}{\text{tan}\ x}=\frac{(\text{tan}\ x)-{x}^{2}}{{x}^{2}\text{tan}\ x}.\]

   As \(x\to {0}^{+},\) the numerator \(\text{tan}\ x-{x}^{2}\to 0\) and the denominator \({x}^{2}\text{tan}\ x\to 0.\) Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have

   \[\underset{x\to {0}^{+}}{\text{lim}}\frac{(\text{tan}\ x)-{x}^{2}}{{x}^{2}\text{tan}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{({\text{sec}}^{2}x)-2x}{{x}^{2}{\text{sec}}^{2}x+2x\ \text{tan}\ x}.\]

   As \(x\to {0}^{+},\) \(({\text{sec}}^{2}x)-2x\to 1\) and \({x}^{2}{\text{sec}}^{2}x+2x\ \text{tan}\ x\to 0.\) Since the denominator is positive as \(x\) approaches zero from the right, we conclude that

   \[\underset{x\to {0}^{+}}{\text{lim}}\frac{({\text{sec}}^{2}x)-2x}{{x}^{2}{\text{sec}}^{2}x+2x\ \text{tan}\ x}=\infty .\]

   Therefore,

   \[\underset{x\to {0}^{+}}{\text{lim}}(\frac{1}{{x}^{2}}-\frac{1}{\text{tan}\ x})=\infty .\]

10. Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}(\frac{1}{x}-\frac{1}{\text{sin}\ x}).\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(0\)

11. Evaluate \(\underset{x\to \infty }{\text{lim}}{x}^{1\text{/}x}.\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    Let \(y={x}^{1\text{/}x}.\) Then,

    \[\text{ln}y=\text{ln}({x}^{1\text{/}x})=\frac{1}{x}\ \text{ln}\ x=\frac{\text{ln}\ x}{x}.\]

    We need to evaluate \(\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{x}.\) Applying L’Hôpital’s rule, we obtain

    \[\underset{x\to \infty }{\text{lim}}\text{ln}\ y=\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{x}=\underset{x\to \infty }{\text{lim}}\frac{1\text{/}x}{1}=0.\]

    Therefore, \(\underset{x\to \infty }{\text{lim}}\text{ln}\ y=0.\) Since the natural logarithm function is continuous, we conclude that

    \[\text{ln}(\underset{x\to \infty }{\text{lim}}y)=0,\]

    which leads to

    \[\underset{x\to \infty }{\text{lim}}y={e}^{\ln (y)}=\text{lim}{e}^{0}=1.\text{(All limits}x\ge \infty \text{.)}\]

    Hence,

    \[\underset{x\to \infty }{\text{lim}}{x}^{1\text{/}x}=1.\]

12. Evaluate \(\underset{x\to \infty }{\text{lim}}{x}^{1\text{/}\text{ln}(x)}.\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(e\)

13. Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}{x}^{\text{sin}\ x}.\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    Let

    \[y={x}^{\text{sin}\ x}.\]

    Therefore,

    \[\text{ln}\ y=\text{ln}({x}^{\text{sin}\ x})=\text{sin}\ x\ \text{ln}\ x.\]

    We now evaluate \(\underset{x\to {0}^{+}}{\text{lim}}\text{sin}\ x\ \text{ln}\ x.\) Since \(\underset{x\to {0}^{+}}{\text{lim}}\text{sin}\ x=0\) and \(\underset{x\to {0}^{+}}{\text{lim}}\text{ln}\ x=\text{-}\infty ,\) we have the indeterminate form \(0\cdot \infty .\) To apply L’Hôpital’s rule, we need to rewrite \(\text{sin}\ x\ \text{ln}\ x\) as a fraction. We could write

    \[\text{sin}\ x\ \text{ln}\ x=\frac{\text{sin}\ x}{1\text{/}\text{ln}\ x}\]

    or

    \[\text{sin}\ x\ \text{ln}\ x=\frac{\text{ln}\ x}{1\text{/}\text{sin}\ x}=\frac{\text{ln}\ x}{\text{csc}\ x}.\]

    Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain

    \[\underset{x\to {0}^{+}}{\text{lim}}\text{sin}\ x\ \text{ln}\ x=\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{sin}\ x}{1\text{/}\text{ln}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{cos}\ x}{-1\text{/}(x{(\text{ln}\ x)}^{2})}=\underset{x\to {0}^{+}}{\text{lim}}(\text{-}x{(\text{ln}\ x)}^{2}\text{cos}\ x).\]

    Unfortunately, we not only have another expression involving the indeterminate form \(0\cdot \infty ,\) but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing

    \[\text{sin}\ x\ \text{ln}\ x=\frac{\text{ln}\ x}{1\text{/}\text{sin}\ x}=\frac{\text{ln}\ x}{\text{csc}\ x},\]

    and applying L’Hôpital’s rule, we obtain

    \[\underset{x\to {0}^{+}}{\text{lim}}\text{sin}\ x\ \text{ln}\ x=\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{ln}\ x}{\text{csc}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{1\text{/}x}{\text{-}\text{csc}\ x\ \text{cot}\ x}=\underset{x\to {0}^{+}}{\text{lim}}\frac{-1}{x\ \text{csc}\ x\ \text{cot}\ x}.\]

    Using the fact that \(\text{csc}\ x=\frac{1}{\text{sin}\ x}\) and \(\text{cot}\ x=\frac{\text{cos}\ x}{\text{sin}\ x},\) we can rewrite the expression on the right-hand side as

    \[\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{-}{\text{sin}}^{2}x}{x\ \text{cos}\ x}=\underset{x\to {0}^{+}}{\text{lim}}[\frac{\text{sin}\ x}{x}\cdot (\text{-}\text{tan}\ x)]=(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{sin}\ x}{x})\cdot (\underset{x\to {0}^{+}}{\text{lim}}(\text{-}\text{tan}\ x))=1\cdot 0=0.\]

    We conclude that \(\underset{x\to {0}^{+}}{\text{lim}}\text{ln}\ y=0.\) Therefore, \(\text{ln}(\underset{x\to {0}^{+}}{\text{lim}}y)=0\) and we have

    \[\underset{x\to {0}^{+}}{\text{lim}}y=\underset{x\to {0}^{+}}{\text{lim}}{x}^{\text{sin}\ x}={e}^{0}=1.\]

    Hence,

    \[\underset{x\to {0}^{+}}{\text{lim}}{x}^{\text{sin}\ x}=1.\]

14. Evaluate \(\underset{x\to {0}^{+}}{\text{lim}}{x}^{x}.\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(1\)

15. For each of the following pairs of functions, use L’Hôpital’s rule to evaluate \(\underset{x\to \infty }{\text{lim}}(\frac{f(x)}{g(x)}).\)

    1. \(f(x)={x}^{2}\ \text{and}\ g(x)={e}^{x}\)
    2. \(f(x)=\text{ln}(x)\ \text{and}\ g(x)={x}^{2}\)
    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    1. Since \(\underset{x\to \infty }{\text{lim}}{x}^{2}=\infty\) and \(\underset{x\to \infty }{\text{lim}}{e}^{x}=\infty ,\) we can use L’Hôpital’s rule to evaluate \(\underset{x\to \infty }{\text{lim}}[\frac{{x}^{2}}{{e}^{x}}].\) We obtain
       \[\underset{x\to \infty }{\text{lim}}\frac{{x}^{2}}{{e}^{x}}=\underset{x\to \infty }{\text{lim}}\frac{2x}{{e}^{x}}.\]
       Since \(\underset{x\to \infty }{\text{lim}}2x=\infty\) and \(\underset{x\to \infty }{\text{lim}}{e}^{x}=\infty ,\) we can apply L’Hôpital’s rule again. Since
       \[\underset{x\to \infty }{\text{lim}}\frac{2x}{{e}^{x}}=\underset{x\to \infty }{\text{lim}}\frac{2}{{e}^{x}}=0,\]
       we conclude that
       \[\underset{x\to \infty }{\text{lim}}\frac{{x}^{2}}{{e}^{x}}=0.\]
       Therefore, \({e}^{x}\) grows more rapidly than \({x}^{2}\) as \(x\to \infty\) (See and ).
       

       \(x\)	\(5\)	\(10\)	\(15\)	\(20\)
       \({x}^{2}\)	\(25\)	\(100\)	\(225\)	\(400\)
       \({e}^{x}\)	\(148\)	\(22,026\)	\(3,269,017\)	\(485,165,195\)

    2. Since \(\underset{x\to \infty }{\text{lim}}\text{ln}\ x=\infty\) and \(\underset{x\to \infty }{\text{lim}}{x}^{2}=\infty ,\) we can use L’Hôpital’s rule to evaluate \(\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{{x}^{2}}.\) We obtain
       \[\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{{x}^{2}}=\underset{x\to \infty }{\text{lim}}\frac{1\text{/}x}{2x}=\underset{x\to \infty }{\text{lim}}\frac{1}{2{x}^{2}}=0.\]
       Thus, \({x}^{2}\) grows more rapidly than \(\text{ln}\ x\) as \(x\to \infty\) (see and ).
       

       \(x\)	\(10\)	\(100\)	\(1000\)	\(10,000\)
       \(\text{ln}(x)\)	\(2.303\)	\(4.605\)	\(6.908\)	\(9.210\)
       \({x}^{2}\)	\(100\)	\(10,000\)	\(1,000,000\)	\(100,000,000\)

16. Compare the growth rates of \({x}^{100}\) and \({2}^{x}.\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    The function \({2}^{x}\) grows faster than \({x}^{100}.\)

17. Evaluate the limit \(\underset{x\to \infty }{\text{lim}}\frac{{e}^{x}}{x}.\)

18. Evaluate the limit \(\underset{x\to \infty }{\text{lim}}\frac{{e}^{x}}{{x}^{k}}.\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(\infty\)

19. Evaluate the limit \(\underset{x\to \infty }{\text{lim}}\frac{\text{ln}\ x}{{x}^{k}}.\)

20. Evaluate the limit \(\underset{x\to a}{\text{lim}}\frac{x-a}{{x}^{2}-{a}^{2}},\ a\ne 0\).

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(\frac{1}{2a}\)

21. Evaluate the limit \(\underset{x\to a}{\text{lim}}\frac{x-a}{{x}^{3}-{a}^{3}},\ a\ne 0\).

22. Evaluate the limit \(\underset{x\to a}{\text{lim}}\frac{x-a}{{x}^{n}-{a}^{n}},\ a\ne 0\).

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(\frac{1}{n{a}^{n-1}}\)

23. \(\underset{x\to {0}^{+}}{\text{lim}}{x}^{2}\text{ln}\ x\)

24. \(\underset{x\to \infty }{\text{lim}}{x}^{1\text{/}x}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    Cannot apply directly; use logarithms

25. \(\underset{x\to 0}{\text{lim}}{x}^{2\text{/}x}\)

26. \(\underset{x\to 0}{\text{lim}}\frac{{x}^{2}}{1\text{/}x}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    Cannot apply directly; rewrite as \(\underset{x\to 0}{\text{lim}}{x}^{3}\)

27. \(\underset{x\to \infty }{\text{lim}}\frac{{e}^{x}}{x}\)

28. \(\underset{x\to 3}{\text{lim}}\frac{{x}^{2}-9}{x-3}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(6\)

29. \(\underset{x\to 3}{\text{lim}}\frac{{x}^{2}-9}{x+3}\)

30. \(\underset{x\to 0}{\text{lim}}\frac{{(1+x)}^{-2}-1}{x}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(-2\)

31. \(\underset{x\to \pi \text{/}2}{\text{lim}}\frac{\text{cos}\ x}{\frac{\pi }{2}-x}\)

32. \(\underset{x\to \pi }{\text{lim}}\frac{x-\pi }{\text{sin}\ x}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(-1\)

33. \(\underset{x\to 1}{\text{lim}}\frac{x-1}{\text{sin}\ x}\)

34. \(\underset{x\to 0}{\text{lim}}\frac{{(1+x)}^{n}-1}{x}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(n\)

35. \(\underset{x\to 0}{\text{lim}}\frac{{(1+x)}^{n}-1-nx}{{x}^{2}}\)

36. \(\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x-\text{tan}\ x}{{x}^{3}}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(-\frac{1}{2}\)

37. \(\underset{x\to 0}{\text{lim}}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\)

38. \(\underset{x\to 0}{\text{lim}}\frac{{e}^{x}-x-1}{{x}^{2}}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(\frac{1}{2}\)

39. \(\underset{x\to {0}^{+}}{\text{lim}}\frac{\text{tan}\ x}{\sqrt{x}}\)

40. \(\underset{x\to 1}{\text{lim}}\frac{x-1}{\text{ln}\ x}\)

    ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

    \(1\)

ନିଜେ ଚେଷ୍ଟାକରନ୍ତୁ

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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