<s>[INST] Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498) 
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's rule, that is, 
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x} = \lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 }$$
since $\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \frac {1 - 1}{ 1- 1} = \frac00$. Thus, we use the L'Hopital's rule again. that is, 
$$\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \lim_{x\rightarrow 0}\frac{-\sin x}{2 \sec^2 x \sec x \tan x}$$
since $\lim_{x\rightarrow 0} \frac{-\sin x}{2 \sec^2 x \sec x \tan x}= \frac{0}{2(1)(0)} = \frac00$. Thus, it always goes to zero by zero. But the answer for this question is $\frac{-1}2$. How is that?!! there must be something missing that either I forget or misunderstand. 
 [/INST] You can make it a little simpler by rewriting
$$\frac{\sin(x)-x}{\tan(x)-x}=\cos(x)\frac{\sin(x)-x}{\sin(x)-x\cos(x)}.$$
The first factor tends to $1$ and can be ignored. Then, applying L'Hospital three times,
$$\frac{\sin(x)-x}{\sin(x)-x\cos(x)}\to\frac{\cos(x)-1}{\cos(x)-\cos(x)+x\sin(x)}\to\frac{-\sin(x)}{\sin(x)+x\cos(x)}\to\frac{-\cos(x)}{\cos(x)+\cos(x)-x\sin(x)}\to-\frac12.$$

Alternatively,
$$\frac{\cos(x)-1}{x\sin(x)}=\frac{-2\sin^2\left(\dfrac x2\right)}{4\left(\dfrac{x}2\right)^2}\frac x{\sin(x)}\to-\frac24\cdot1^2\cdot1.$$
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