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Mathematics LibreTexts

3.2: L'Hôpital's Rule

Proof of L'Hôpital's Rule

Goal:  Easily find 

\[ \lim _{x \rightarrow 0} \dfrac{\sin\, x}{x}.\]

Suppose that \(f\) and \(g\) are continuous functions and 

\[ f(a) = g(a) = 0.\]

Recall the mean value theorem states that

\[ f'(c) = \dfrac{f(b)-f(a)}{b-a} \]

so that

\[ \dfrac{ f'(c)}{g'(c)} = \dfrac{\dfrac{f(b)-f(a)}{b-a}}{\dfrac{g(b)-g(a)}{b-a}} = \dfrac{f(b)-f(a)}{g(b)-g(a)}.\]

Let

\[a = 0.\]

Then 

\[ \dfrac{f'(c)}{g'(c)} = \dfrac{f(b)-f(0)}{g(b)-g(0)} = \dfrac{f(b)-0}{g(b)-0} =\dfrac{f(b)}{g(b)}\]

so that

\[ \lim_{b \rightarrow 0} \dfrac{f(b)}{g(b)} = \lim_{c \rightarrow 0} \dfrac{f'(c)}{g'(c)}.\]

Hence

\[ \lim_{x \rightarrow 0} \dfrac{\sin\,x}{x} = \lim_{x \rightarrow 0} \dfrac{\cos \, x}{1} = 1.\]

Definition: L'Hôpital's Rule

Let

\[ f(c) = g(c) = 0  \]

then

\[ \lim_{x \rightarrow c} \dfrac{f(x)}{g(x)} = \lim_{x \rightarrow c} \dfrac{f'(x)}{g'(x)}. \nonumber\]

Example 1

\[ \lim_{x \rightarrow 0} \dfrac{e^x - 1}{x} =  \lim_{x \rightarrow 0} \dfrac{e^x }{1} = 1. \nonumber\]

Exercise

Exercises:  Determine the following limits if they exist

  1. \( \lim_{x \rightarrow 1} \dfrac{\ln x}{x^2 -1}\)
  2. \( \lim_{x \rightarrow 0} \dfrac{\sin x}{x + 1}\)
  3. \( \lim_{x \rightarrow \infty} \dfrac{e^{-x}}{x^2}\).

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 \( \infty - \infty \).

Example 1: \(0\times\infty\)

\[\begin{align} \lim_{x\to{\infty}}\big(\arctan x -\dfrac{\pi}{2}e^x\big) &= \lim_{x\to{\infty}} \Big(\dfrac{\arctan x -\dfrac{\pi}{2}}{e^{-x}}\Big) \\ &=\lim_{x\to{\infty}} \dfrac{\dfrac{1}{1+x^2}}{-e^{-x}} \\ &= -\lim_{x\to{\infty}}\dfrac{e^x}{1+x^2} \\ &= -\lim_{x\to{\infty}} \dfrac{e^x}{2x} \\ &= -\lim_{x\to{\infty}}\dfrac{e^x}{2} \\&= -\infty. \end{align}\]

Example 2:  \((\infty)^0\)

\[\begin{align} \lim_{x\to{\infty}}\big(1+\dfrac{1}{x}\big)^x &= e^{ \lim_{x\to{\infty}}x\ln(1+\frac{1}{x})} \\ &=   e^{ \lim_{x\to{\infty}} \frac{ \ln(1+\frac{1}{x})}{\frac{1}{x}}} \\ &=  e^{ \lim_{x\to{\infty}} \frac{ -\frac{1}{x^2}\big(\frac{1}{1+\frac{1}{x}}\big)}{-\frac{1}{x^2}}} \\ &= e^{\lim_{x\to{\infty}} \frac{1}{1+\frac{1}{x}}} \\ &= e^1 \\ &= e.   \end{align}\]

Example 3: \(\infty - \infty\)

\[\begin{align} \lim_{x\to{1^+}} \Big[\dfrac{1}{x^2-1}-\dfrac{1}{\ln x}\big] &= \lim_{x\to{1^+}}\Big[ \dfrac{\ln x -(x^2-1)}{(x^2-1)\ln x} \Big] \\ &= \lim_{x\to{1^+}}\Big[ \dfrac{\dfrac{1}{x}-2x}{2x\ln x+\dfrac{x^2-1}{x}} \Big] \\ &= \dfrac{1-2}{0}. \end{align}\]

Exercise 1

Evaluate

\[ \lim_{x \rightarrow 0^+} x^x.\]

Contributors

  • Integrated by Justin Marshall.