Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 24

Determine the $\displaystyle \lim_{x \to 0} \frac{x - \sin x}{x - \tan x}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to 0} \frac{x - \sin x}{x - \tan x} = \frac{0- \sin 0}{0 - \tan 0} = \frac{0}{0} \text{ Indeterminate form}$
Thus, by applying L'Hospital's Rule...
$\displaystyle \lim_{x \to 0} \frac{x - \sin x}{x - \tan x} = \lim_{x \ to 0} \frac{1 - \cos x}{1 - \sec^2 x}$

We will still get indeterminate form by evaluating the limit. So we will apply L'Hospital's Rule once more,

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