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. 

Notice that $\tan x = \sin x \sec x$; so we'll have:
$$\lim_{x \to 0} \frac{-\sin x}{2\sec^2x\sec x \tan x} = \lim_{x \to 0} \frac{-\sin x}{2\sec^2x\sec^2 x \sin x} = \lim_{x \to 0} \frac{-\sin x}{2\sec^4x\sin x} = \lim_{x \to 0} \frac{-1}{2\sec^4x} = \cdots$$
Hope you can go from here. :x
