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. 
 A: $$\frac{\cos(x)-1}{\sec^2(x)-1}=\frac{\cos(x)-1}{\frac{1}{\cos^2(x)}-1}=\frac{\cos^2(x)(\cos(x)-1))}{1-\cos^2(x)}=\frac{\cos^2(x)(\cos(x)-1)}{(1-\cos(x))(1+\cos(x))}=\frac{- \cos^2(x)(1-\cos(x))}{(1-\cos(x))(1+\cos(x))}$$
A: 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
A: Hint: use the limit development of sin and tan
A: Hint differentiate again you get $$\frac{-\cos(x)}{2(\sec^4(x)+2\tan^2(x)\sec^2(x))}$$ now whats the limit
A: For $x\to 0$ we have $\sin x\sim x-\frac{x^3}{3!} $ and $\tan x\sim x+\frac{x^3}{3} $
then 
$$
\frac{\sin x- x}{\tan x-x}\sim \frac{-\frac{x^3}{3!}}{+\frac{x^3}{3}}=-\frac{1}{2}
$$
A: $$
\lim_{x\to0}\frac{\sin x-x}{\tan x-x}\overset{L'H}{=}\lim_{x\to0}\frac{\cos x-1}{\frac{1}{\cos^2x}-1}=\lim_{x\to0}\frac{\frac{\cos x-1}{x^2}}{\frac{1-\cos^2 x}{x^2\cos^2x}}
$$
now, 
$$
\lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac{1}{2}
$$ 
and
$$
\lim_{x\to0}\frac{1-\cos^2 x}{x^2\cos^2x}=\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\frac{1}{\cos^2x}=1
$$
so, the desired limit is $-\frac{1}{2}$
A: 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,
$$\frac{\cos(x)-1}{x\sin(x)}=\frac{-2\sin^2\left(\dfrac x2\right)}{4\left(\dfrac{x}2\right)^2}\frac x{\sin(x)}\to-\frac24\cdot1^2\cdot1.$$
