7.7E: Exercises for L'Hôpital's Rule

Last updated

Jun 23, 2021
Save as PDF
- 7.7: L'Hôpital's Rule
- 7.8: Improper Integrals

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$ $\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

Terms and Concepts

1. List the different indeterminate forms described in this section.

Answer:: The forms $0⋅∞, ∞−∞, 1^∞, ∞^0$ , and $0^0$ are all considered indeterminate.

2. List similar looking forms that are not indeterminate.

Answer:: Among others, the forms $∞⋅∞, ∞+∞, -∞−∞, 0^∞, 1^0$ , and $∞^∞$ are not considered indeterminate, as these limits can be determined clearly.

3. T/F: l'Hôpital's Rule states that $\frac{d}{dx} \left ( \frac{f(x)}{g(x)}\right ) = \frac{f'(x)}{g'(x)}$ .

Answer:: False. L'Hôpital's Rule is a method for taking limits of rational functions in certain cases. It does not replace the Quotient Rule when taking the derivative of these rational functions.

4. Explain what the indeterminate form " $1^{\infty}$ " means. Why is it indeterminate?

Answer:: When a limit has the form " $1^{\infty}$ ", this means that the function in the base of the exponent is approaching $1$ , while the function in the exponent is approaching $\infty$ . It's indeterminate, since if the base function is approaching 1, but always is less than 1, then the limit could be 0, while if the base function were approaching 1, but always is greater than 1, the limit could be $\infty$ . But since this uncertainty exists, the limit could, in fact, be anything.

5. Explain why limits of the form $\infty - \infty$ are indeterminate.

Answer:: Limits with this form depend on the relative speed with which the two terms are approaching $\infty$ . If the first term approaches $\infty$ faster than the second term, the limit would be $\infty$ . If the second term approaches $\infty$ faster than the first term, the limit would be $-\infty$ . But if they both appraoch $\infty$ at about the same rates, the limit could be anything!

6. Fill in the blanks" The Quotient Rule is applied to $\frac{f(x)}{g(x)}$ when taking its _____; l'Hôpital's Rule is applied when taking _______ of $\frac{f(x)}{g(x)}$ when the form is _______ or _______.

Answer:: derivative; limits; $\dfrac{0}{0}$ or $\dfrac{\pm\infty}{\pm\infty}$

7. Create (but do not evaluate) a limit that initially has the form " $\infty^0$ ".

8. Create a function $f(x)$ such that $\lim\limits_{x\to1}f(x)$ initially has the form " $0^0$ ".

Problems

For exercises 1 - 6, evaluate the limit.

1) Evaluate the limit $\displaystyle \lim_{x→∞}\frac{e^x}{x}$ .

2) Evaluate the limit $\displaystyle \lim_{x→∞}\frac{e^x}{x^k}$ .

Answer:: $\displaystyle \lim_{x→∞}\frac{e^x}{x^k} \quad = \quad ∞$

3) Evaluate the limit $\displaystyle \lim_{x→∞}\frac{\ln x}{x^k}$ .

4) Evaluate the limit $\displaystyle \lim_{x→a}\frac{x−a}{x^2−a^2}$ .

Answer:: $\displaystyle \lim_{x→a}\frac{x−a}{x^2−a^2} \quad = \quad \frac{1}{2a}$

5. Evaluate the limit $\displaystyle \lim_{x→a}\frac{x−a}{x^3−a^3}$ .

6. Evaluate the limit $\displaystyle \lim_{x→a}\frac{x−a}{x^n−a^n}$ .

Answer:: $\displaystyle \lim_{x→a}\frac{x−a}{x^n−a^n} \quad = \quad \frac{1}{na^{n−1}}$

For exercises 7 - 11, determine whether you can apply L’Hôpital’s rule directly. Explain why or why not. Then, indicate if there is some way you can alter the limit so you can apply L’Hôpital’s rule.

7) $\displaystyle \lim_{x→0^+}x^2\ln x$

8) $\displaystyle \lim_{x→∞}x^{1/x}$

Answer:: Cannot apply directly; use logarithms

9) $\displaystyle \lim_{x→0}x^{2/x}$

10) $\displaystyle \lim_{x→0}\frac{x^2}{1/x}$

Answer:: Cannot apply directly; rewrite as $\displaystyle \lim_{x→0}x^3$

11) $\displaystyle \lim_{x→∞}\frac{e^x}{x}$

For exercises 12 - 44, evaluate the limits with either L’Hôpital’s rule or previously learned methods.

12) $\displaystyle \lim_{x→3}\frac{x^2−9}{x−3}$

Answer:: $\displaystyle \lim_{x→3}\frac{x^2−9}{x−3} \quad = \quad 6$

13) $\displaystyle \lim_{x→3}\frac{x^2−9}{x+3}$

14) $\displaystyle \lim_{x→0}\frac{(1+x)^{−2}−1}{x}$

Answer:: $\displaystyle \lim_{x→0}\frac{(1+x)^{−2}−1}{x} \quad = \quad -2$

15) $\displaystyle \lim_{x→π/2}\frac{\cos x}{\frac{π}{2}−x}$

16) $\displaystyle \lim_{x→π}\frac{x−π}{\sin x}$

Answer:: $\displaystyle \lim_{x→π}\frac{x−π}{\sin x} \quad = \quad -1$

17) $\displaystyle \lim_{x→1}\frac{x−1}{\sin x}$

18) $\displaystyle \lim_{x→0}\frac{(1+x)^n−1}{x}$

Answer:: $\displaystyle \lim_{x→0}\frac{(1+x)^n−1}{x} \quad = \quad n$

19) $\displaystyle \lim_{x→0}\frac{(1+x)^n−1−nx}{x^2}$

20) $\displaystyle \lim_{x→0}\frac{\sin x−\tan x}{x^3}$

Answer:: $\displaystyle \lim_{x→0}\frac{\sin x−\tan x}{x^3} \quad = \quad −\frac{1}{2}$

21) $\displaystyle \lim_{x→0}\frac{\sqrt{1+x}−\sqrt{1−x}}{x}$

22) $\displaystyle \lim_{x→0}\frac{e^x−x−1}{x^2}$

Answer:: $\displaystyle \lim_{x→0}\frac{e^x−x−1}{x^2} \quad = \quad \frac{1}{2}$

23) $\displaystyle \lim_{x→0}\frac{\tan x}{\sqrt{x}}$

24) $\displaystyle \lim_{x→1}\frac{x-1}{\ln x}$

Answer:: $\displaystyle \lim_{x→1}\frac{x-1}{\ln x} \quad = \quad 1$

25) $\displaystyle \lim_{x→0}\,(x+1)^{1/x}$

26) $\displaystyle \lim_{x→1}\frac{\sqrt{x}−\sqrt[3]{x}}{x−1}$

Answer:: $\displaystyle \lim_{x→1}\frac{\sqrt{x}−\sqrt[3]{x}}{x−1} \quad = \quad \frac{1}{6}$

27) $\displaystyle \lim_{x→0^+}x^{2x}$

28) $\displaystyle \lim_{x→∞}x\sin\left(\tfrac{1}{x}\right)$

Answer:: $\displaystyle \lim_{x→∞}x\sin\left(\tfrac{1}{x}\right) \quad = \quad 1$

29) $\displaystyle \lim_{x→0}\frac{\sin x−x}{x^2}$

30) $\displaystyle \lim_{x→0^+}x\ln\left(x^4\right)$

Answer:: $\displaystyle \lim_{x→0^+}x\ln\left(x^4\right) \quad = \quad 0$

31) $\displaystyle \lim_{x→∞}(x−e^x)$

32) $\displaystyle \lim_{x→∞}x^2e^{−x}$

Answer:: $\displaystyle \lim_{x→∞}x^2e^{−x} \quad = \quad 0$

33) $\displaystyle \lim_{x\to 1^+} \left[\frac{1}{\ln x}-\frac{1}{1-x}\right]$

34) $\displaystyle \lim_{x\to 3^+} \left[\frac{5}{x^2-9}-\frac{x}{x-3}\right]$

Answer:: $\displaystyle \lim_{x\to 3^+} \left[\frac{5}{x^2-9}-\frac{x}{x-3}\right] = \lim_{x\to 3^+} \frac{5-x^2-3x}{x^2-9} \quad = \quad -∞$

35) $\displaystyle \lim_{x\to \infty} \frac{\sqrt{2x^2-3}}{x+2}$

Note:: L’Hôpital’s rule fails to help us find this limit, although the form seems appropriate. But you can evaluate this limit using techniques you learned earlier in calculus.

36) $\displaystyle \lim_{x\to \infty} \left(\frac{x+7}{x+3}\right)^{x}$

Answer:: $\displaystyle \lim_{x\to \infty} \left(\frac{x+7}{x+3}\right)^{x} \quad = \quad e^{4}$

37) $\displaystyle \lim_{x→0}\frac{3^x−2^x}{x}$

38) $\displaystyle \lim_{x→0}\frac{1+1/x}{1−1/x}$

Answer:: $\displaystyle \lim_{x→0}\frac{1+1/x}{1−1/x} \quad = \quad -1$

39) $\displaystyle \lim_{x→π/4}(1−\tan x)\cot x$

40) $\displaystyle \lim_{x→∞}xe^{1/x}$

Answer:: $\displaystyle \lim_{x→∞}xe^{1/x} \quad = \quad ∞$

41) $\displaystyle \lim_{x→0}x^{1/\cos x}$

42) $\displaystyle \lim_{x→0^+}x^{1/x}$

Answer:: $\displaystyle \lim_{x→0^+}x^{1/x} \quad = \quad 0$

43) $\displaystyle \lim_{x→0}\left(1−\frac{1}{x}\right)^x$

44) $\displaystyle \lim_{x→∞}\left(1−\frac{1}{x}\right)^x$

Answer:: $\displaystyle \lim_{x→∞}\left(1−\frac{1}{x}\right)^x \quad = \quad \frac{1}{e}$

For exercises 45 - 54, use a calculator to graph the function and estimate the value of the limit, then use L’Hôpital’s rule to find the limit directly.

45) [T] $\displaystyle \lim_{x→0}\frac{e^x−1}{x}$

46) [T] $\displaystyle \lim_{x→0}x\sin\left(\tfrac{1}{x}\right)$

Answer:: $\displaystyle \lim_{x→0}x\sin\left(\tfrac{1}{x}\right) \quad = \quad 0$

47) [T] $\displaystyle \lim_{x→1}\frac{x−1}{1−\cos(πx)}$

48) [T] $\displaystyle \lim_{x→1}\frac{e^{x−1}−1}{x−1}$

Answer:: $\displaystyle \lim_{x→1}\frac{e^{x−1}−1}{x−1} \quad = \quad 1$

49) [T] $\displaystyle \lim_{x→1}\frac{(x−1)^2}{\ln x}$

50) [T] $\displaystyle \lim_{x→π}\frac{1+\cos x}{\sin x}$

Answer:: $\displaystyle \lim_{x→π}\frac{1+\cos x}{\sin x} \quad = \quad 0$

51) [T] $\displaystyle \lim_{x→0}\left(\csc x−\frac{1}{x}\right)$

52) [T] $\displaystyle \lim_{x→0^+}\tan\left(x^x\right)$

Answer:: $\displaystyle \lim_{x→0^+}\tan\left(x^x\right) \quad = \quad \tan 1$

53) [T] $\displaystyle \lim_{x→0^+}\frac{\ln x}{\sin x}$

54) [T] $\displaystyle \lim_{x→0}\frac{e^x−e^{−x}}{x}$

Answer:: $\displaystyle \lim_{x→0}\frac{e^x−e^{−x}}{x} \quad = \quad 2$

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.
Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/
Terms and Concepts 1-4, and 6-8 were adapted from Apex Calculus by Paul Seeburger (Monroe Community College)
Terms and Concepts Problem 5 was created by Paul Seeburger
Problems 33 and 34 from Apex Calculus. Solution to 34 by Paul Seeburger
Problems 35 and 36 and solutions for Terms and Concepts Problems 1-6 and for Problem 36 were added by Paul Seeburger

Search

Text Color

Text Size

Margin Size

Font Type

Terms and Concepts

Problems

Contributors

Support Center

How can we help?