Find $\lim_{x\to 0}\frac{x}{\sin x - x}$ Find limit:
$$\lim_{x\to 0}\frac{x}{\sin x - x} = ???$$
Any help would be thoroughly appreciated.
 A: Recall that $\sin(x)=x-\dfrac{x^3}{3!}+o(x^3)$. Then
$$\lim_{x\to 0} \frac {x}{\sin(x)-x} = \lim_{x\to 0} \frac {x}{-\frac{x^3}{6} + o(x^3)}= \lim_{x\to 0} \frac {-6}{x^2 + o(x^2)} = -\infty$$
A: An idea: you can write
$$\frac x{\sin x-x}=\frac1{\frac{\sin x}x-1}$$
and use now that $\dfrac{\sin x}x\xrightarrow[x\to 0]{}1\;$ to show the limit doesn't exist finitely.
A: You should know that $\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$. 
Then, look at $$\lim_{x \to 0} \frac{\sin x - x}{x} = \lim_{x \to 0} \frac{\sin x}{x} - \lim_{x \to 0} \frac{x}{x} = 1 - \lim_{x \to 0} 1 = 1 - 1  = 0.$$ 
So, the limit you have doesn't exist (since it would behave like the reciprocal of teh limit I just stated.
A: If you're allowed to use l’Hôpital's theorem, you have
$$
\lim_{x\to0}\frac{x}{\sin x-x}=\lim_{x\to0}\frac{1}{\cos x-1}
$$
Since $\cos x-1<0$ for $-\pi/2<x<\pi/2$, $x\ne0$, we know that the denominator is negative in a (punctured) neighborhood of $0$. Thus
$$
\lim_{x\to0}\frac{x}{\sin x-x}=\lim_{x\to0}\frac{1}{\cos x-1}=-\infty
$$
