Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 68

Illustrate L'Hospital's Rule by graphing both $\displaystyle \frac{f(x)}{g(x)}$ and $\displaystyle \frac{f'(x)}{g'(x)}$ near $x = 0$ to see that these ratios have the same limit $x \to 0$. Alos calculate the exact value of the limit.


If $f(x) = 2 x \sin x$, then by using product rule...

$
\begin{equation}
\begin{aligned}
f'(x) &= 2 [ x(\cos x) + (1) \sin x]\\
\\
f'(x) &= 2x \cos x + 2 \sin x\\
\\
\text{Also, if } g(x) &= \sec x - 1 \text{ ,then}\\
\\
g'(x) &= \sec x \tan x
\end{aligned}
\end{equation}
$

So,
$\displaystyle \frac{f(x)}{g(x)} = \frac{2x \sin x}{\sec x - 1} \text{ and } \frac{f'(x)}{g'(x)} = \frac{2x \cos x + 2 \sin x}{\sec x \tan x}$




Based from the graph, the values of the $\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)} \approx \lim_{x \to 0} \frac{f'(x)}{g'(x)} \approx 4$
To find the exact value, we will use L'Hospital's Rule

For $\displaystyle \lim_{x \to 0} \frac{f(x)}{g(x)}$,


$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{2x \sin x}{\sec x - 1} &= \lim_{x \to 0} \frac{\frac{d}{dx} (2x \sin x)}{\frac{d}{dx}(\sec x -1) }\\
\\
&= \lim_{x \to 0} \frac{2x \cos x + 2 \sin x}{\sec x \tan x}
\end{aligned}
\end{equation}
$


If we evaluate the limit, we will still get an indeterminate form, so we must use apply the L'Hospital's Rule once more...
$ $


$
\begin{equation}
\begin{aligned}
&= \lim_{x \to 0} \frac{2[x (-\sin x) + (1) \cos x] + 2 \cos x}{\sec x (\sec^2 x) + (\sec x \tan x)(\tan x)}\\
\\
&= \lim_{x \to 0} \frac{-2x \sin x + 4 \cos x}{\sec^3 x + \sec x \tan^2 x}\\
\\
&= \frac{-2(0)\sin(0)+4\cos(0)}{\sec^3(0) + \sec (0) \tan^2 (0)}\\
\\
&= \frac{0 + 4(1)}{1+0}\\
\\
&= \frac{4}{1}\\
\\
&= 4
\end{aligned}
\end{equation}
$