Home / AP Calculus AB and BC: Chapter 6 – Techniques of Integration : 6.4 – L’Hospital’s Rule Study Notes

AP Calculus AB and BC: Chapter 6 – Techniques of Integration : 6.4 – L’Hospital’s Rule Study Notes

 L’Hospital’s Rule

L’Hospital’s Rule
Suppose $f$ and $g$ are differentiable and $g^{\prime}(x) \neq 0$ near $x=c$ (except possibly at $c$ ).
If the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches $c$ produces the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then

$
\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}
$
provided the limit on the right side exists.
L’Hospital’s Rule can be applied only to quotients leading to indeterminate forms such as

$\frac{0}{0}, \frac{\infty}{\infty}, 0 \cdot \infty, 1^{\infty}, \infty^0, 0^0$, and $\infty-\infty$.

L’Hospital’s Rule does not apply when either the numerator or denominator has a finite nonzero limit.

Example 1

  • Find $\lim _{x \rightarrow 0} \frac{e^x-1}{\sin x}$
▶️Answer/Explanation

Solution

Indeterminate form $\frac{0}{0}$
L’Hospital’s Rule: $\frac{d}{d x}\left(e^x-1\right)=e^x, \frac{d}{d x}(\sin x)=\cos x$. $e^0=1$ and $\cos 0=1$

$e^0=1$ and $\cos 0=1$.

$\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^x-1}{\sin x} \\ & =\lim _{x \rightarrow 0} \frac{e^x}{\cos x} \\ & =1\end{aligned}$

Example 2

  • Find $\lim _{x \rightarrow \pi / 2} \frac{\sec x+9}{\tan x}$.
▶️Answer/Explanation

Solution

$\lim _{x \rightarrow \pi / 2} \frac{\sec x+9}{\tan x}$ Indeterminate form $\frac{\infty}{\infty}$.

$=\lim _{x \rightarrow \pi / 2} \frac{\sec x \tan x}{\sec ^2 x}$                    L’Hospital’s Rule: $\frac{d}{d x}(\sec x+9)=\sec x \tan x$,

$\frac{d}{d x}(\tan x)=\sec ^2 x$

$=\lim _{x \rightarrow \pi / 2} \frac{\tan x}{\sec x}$                                                                  Simplify

$\begin{aligned} & =\lim _{x \rightarrow \pi / 2} \sin x \\ & =1\end{aligned}$

Example 3

  • Find $\lim _{x \rightarrow \infty} x \tan \frac{1}{x}$.
▶️Answer/Explanation

Solution

$
\begin{array}{ll}
\lim _{x \rightarrow 0} \frac{e^x-1}{\sin x} & \text { Indeterminate form } \frac{0}{0} \\
=\lim _{x \rightarrow 0} \frac{e^x}{\cos x} & \text { L’Hospital’s Rule: } \frac{d}{d x}\left(e^x-1\right)=e^x, \frac{d}{d x}(\sin x)=\cos x . \\
=1 & e^0=1 \text { and } \cos 0=1 .
\end{array}
$

Example4

  • $
    \lim _{x \rightarrow 0} \frac{e^x-1-x}{x^2}=
    $

(A) 0

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

(C) 1

(D) $\infty$

▶️Answer/Explanation

Ans:B

Example5

  • $
    \lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=
    $

(A) $-\infty$

(B) 0

(C) $\frac{\pi}{2}$

(D) 1

▶️Answer/Explanation

Ans:D

Example 6

$
\lim _{\theta \rightarrow \pi} \frac{\sin \theta}{\theta-\pi}=
$

(A) -1

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

(C) 0

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

▶️Answer/Explanation

Ans:A

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