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:
$$\begin{align}
&\lim_{x\rightarrow 0} \frac{\cos x - 1}{x} \\
-1 \le &\lim_{x\rightarrow 0} (\cos x)(x^{-1}) \le 1 \\
\lim_{x\rightarrow 0} -x^{-1} \le &\lim_{x\rightarrow 0} \cos x \le \lim_{x\rightarrow 0} x^{-1} \quad \text{*}
\end{align}$$
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. 
 A: Hint 1: $$\cos x -1 = -2\sin^2 \dfrac{x}2$$
Hint 2: $$\dfrac{\cos x - 1}{x} = \dfrac{(\cos x - 1)(\cos x + 1)}{x(\cos x + 1)} = -\dfrac{\sin^2 x}{x^2} \dfrac{x}{1+\cos x}.$$
A: If you know that $\sin \theta \lt \theta$ when $\theta\gt0$, then you can argue that
$$0\lt\left| {1-\cos x\over x}\right|\lt\left|{1-\cos^2x\over x}\right|=\left|{\sin^2x\over x}\right|\lt \left|\sin x\right|$$
at which point the Squeeze Theorem takes over.  If you need to prove the inequality $\sin\theta\lt\theta$, note that ${1\over2}\sin\theta$ (for small $\theta$, at least) is the area of the triangle inside the unit circle with vertices at the origin, $(1,0)$, and $(\cos\theta,\sin\theta)$, whereas ${1\over2}\theta$ is the area of the sector containing the triangle. (Alternatively, note that $\sin\theta$, being the vertical distance from $(\cos\theta,\sin\theta)$ to the $x$-axis, is less than the distance from $(\cos\theta,\sin\theta)$ to $(1,0)$, which is less than the arc length connecting those two points, namely $\theta$.)
A: $\newcommand{\+}{^{\dagger}}%
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By Mean Value Theorem:
$$
\verts{\cos\pars{x} - 1 \over x} = \verts{\sin\pars{\xi}}
\quad\mbox{where}\quad
\left\vert%
\begin{array}{lcl}
\quad 0 < \xi < x & \mbox{if} & x > 0
\\
\mbox{or}&&
\\
\quad x < \xi < 0 & \mbox{if} & x < 0
\end{array}\right.
$$
Now, use 'Sandwich Theorem'.
A: Let $f(x) = \cos(x)$, then $\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} = f'(0) = -\sin(0) = 0$
A: It is sufficient to use the asymptotic development of $\cos x$ around $0$, namely
$$\cos x = 1 - \frac{x^2}{2} + o(x^3).$$
