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...



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