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# 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 $$\PageIndex{1}$$

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

Exercise $$\PageIndex{1}$$

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 $$\PageIndex{2}$$ : $$0\times\infty$$

\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 $$\PageIndex{3}$$ : $$(\infty)^0$$

\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 $$\PageIndex{3}$$: $$\infty - \infty$$

\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 $$\PageIndex{2}$$

Evaluate

$\lim_{x \rightarrow 0^+} x^x.$

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.