Find $\lim_{x\rightarrow 0} \frac{\cos x - 1}{x}$ I'm trying to find the following limit: 

$$\lim_{x\rightarrow 0} \frac{\cos x - 1}{x}$$

I tried to use squeeze theorem but it's not making much sense. I did the following:

$$\begin{align}
&amp;\lim_{x\rightarrow 0} \frac{\cos x - 1}{x} \\
-1 \le &amp;\lim_{x\rightarrow 0} (\cos x)(x^{-1}) \le 1 \\
\lim_{x\rightarrow 0} -x^{-1} \le &amp;\lim_{x\rightarrow 0} \cos x \le \lim_{x\rightarrow 0} x^{-1} \quad \text{*}
\end{align}$$

The last line is where I'm confused. I don't think I'm doing squeeze theorem correctly. I'm guessing you have to manipulate $\cos x - 1$ somehow. Please provide some hints.

Thanks a bunch!

P.S You cannot use L'Hopital.
Hint 1: $$\cos x -1 = -2\sin^2 \dfrac{x}2$$
Hint 2: $$\dfrac{\cos x - 1}{x} = \dfrac{(\cos x - 1)(\cos x + 1)}{x(\cos x + 1)} = -\dfrac{\sin^2 x}{x^2} \dfrac{x}{1+\cos x}.$$
It is sufficient to use the asymptotic development of $\cos x$ around $0$, namely

$$\cos x = 1 - \frac{x^2}{2} + o(x^3).$$