3.2: L'Hôpital's Rule
- Page ID
- 495
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Proof of L'Hôpital's Rule
Goal: Easily find
\[ \lim _{x \rightarrow 0} \dfrac{\sin\, x}{x}.\nonumber \]
Suppose that \(f\) and \(g\) are continuous functions and
\[ f(a) = g(a) = 0.\nonumber \]
Recall the mean value theorem states that
\[ f'(c) = \dfrac{f(b)-f(a)}{b-a} \nonumber \]
so that
\[ \dfrac{ f'(c)}{g'(c)} = \dfrac{\dfrac{f(b)-f(a)}{b-a}}{\dfrac{g(b)-g(a)}{b-a}} = \dfrac{f(b)-f(a)}{g(b)-g(a)}.\nonumber \]
Let
\[a = 0.\nonumber \]
Then
\[ \dfrac{f'(c)}{g'(c)} = \dfrac{f(b)-f(0)}{g(b)-g(0)} = \dfrac{f(b)-0}{g(b)-0} =\dfrac{f(b)}{g(b)}\nonumber \]
so that
\[ \lim_{b \rightarrow 0} \dfrac{f(b)}{g(b)} = \lim_{c \rightarrow 0} \dfrac{f'(c)}{g'(c)}.\nonumber \]
Hence
\[ \lim_{x \rightarrow 0} \dfrac{\sin\,x}{x} = \lim_{x \rightarrow 0} \dfrac{\cos \, x}{1} = 1.\nonumber \]
Let
\[ f(c) = g(c) = 0 \nonumber \]
then
\[ \lim_{x \rightarrow c} \dfrac{f(x)}{g(x)} = \lim_{x \rightarrow c} \dfrac{f'(x)}{g'(x)}. \nonumber \nonumber \]
\[ \lim_{x \rightarrow 0} \dfrac{e^x - 1}{x} = \lim_{x \rightarrow 0} \dfrac{e^x }{1} = 1. \nonumber \nonumber \]
Determine the following limits if they exist
- \( \lim_{x \rightarrow 1} \dfrac{\ln x}{x^2 -1}\)
- \( \lim_{x \rightarrow 0} \dfrac{\sin x}{x + 1}\)
- \( \lim_{x \rightarrow \infty} \dfrac{e^{-x}}{x^2}\).
Hidden Forms of L'Hôpital's Rule
We can also use L'Hôpital's rule when we have expressions of the form
\( 0 \times \infty\), \( \infty^0\), and \( \infty - \infty \).
\[\begin{align*} \lim_{x\to{\infty}}\big(\arctan x -\dfrac{\pi}{2}e^x\big) &= \lim_{x\to{\infty}} \Big(\dfrac{\arctan x -\dfrac{\pi}{2}}{e^{-x}}\Big) \\[4pt] &=\lim_{x\to{\infty}} \dfrac{\dfrac{1}{1+x^2}}{-e^{-x}} \\[4pt] &= -\lim_{x\to{\infty}}\dfrac{e^x}{1+x^2} \\[4pt] &= -\lim_{x\to{\infty}} \dfrac{e^x}{2x} \\[4pt] &= -\lim_{x\to{\infty}}\dfrac{e^x}{2} \\[4pt]&= -\infty. \end{align*}\nonumber \]
\[\begin{align*} \lim_{x\to{\infty}}\big(1+\dfrac{1}{x}\big)^x &= e^{ \lim_{x\to{\infty}}x\ln(1+\frac{1}{x})} \\[4pt] &= e^{ \lim_{x\to{\infty}} \frac{ \ln(1+\frac{1}{x})}{\frac{1}{x}}} \\[4pt] &= e^{ \lim_{x\to{\infty}} \frac{ -\frac{1}{x^2}\big(\frac{1}{1+\frac{1}{x}}\big)}{-\frac{1}{x^2}}} \\[4pt] &= e^{\lim_{x\to{\infty}} \frac{1}{1+\frac{1}{x}}} \\[4pt] &= e^1 \\[4pt] &= e. \end{align*}\nonumber \]
\[\begin{align*} \lim_{x\to{1^+}} \Big[\dfrac{1}{x^2-1}-\dfrac{1}{\ln x}\big] &= \lim_{x\to{1^+}}\Big[ \dfrac{\ln x -(x^2-1)}{(x^2-1)\ln x} \Big] \\[4pt] &= \lim_{x\to{1^+}}\Big[ \dfrac{\dfrac{1}{x}-2x}{2x\ln x+\dfrac{x^2-1}{x}} \Big] \\[4pt] &= \dfrac{1-2}{0}. \end{align*}\nonumber \]
Evaluate
\[ \lim_{x \rightarrow 0^+} x^x.\nonumber \]
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.