Finding $\lim \limits_{x \to 0} \frac{1 - \cos x}{x}$, given $\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$ I am working on a textbook problem. The first step is to prove that
$$\lim \limits_{x \to 0} \frac{\sin x}{x} = 1$$
(which I did). The exercise goes on

Use this limit [i.e. the one above] to find 
  $$\lim \limits_{x \to 0} \frac{1 - \cos x}{x}$$.

How can this be done? I don't really see a connection between the two...
 A: $$
\begin{eqnarray*} 
\lim_{x \to 0} \frac{1 - \cos{x}}{x}  
&=& \lim_{x \to 0} \frac{(1-\cos{x})(1+\cos x)}{x(1+\cos x)} 
\\ &=& \lim_{x \to 0} \frac{1-\cos^2 x}{x(1 + \cos x)} 
\\ &=& \lim_{x \to 0} \frac{x\sin^2 x}{x \cdot x(1+ \cos x)} 
\\ &=& \lim_{x \to 0} \frac{\sin x}{x} \times \frac{\sin{x}}{x}\times \frac{x}{1+\cos x} = 0 .
\end{eqnarray*}
$$
A: \begin{align*}
L = \lim_{x \to 0} \frac{1-\cos{x}}{x} &= \lim_{x \to 0} \frac{\sin^{2}{x} + \cos^{2}{x} -\cos{x}}{x} \\ &= \lim_{x \to 0} \cos(x) \cdot \frac{\cos{x}-1}{x} + \lim_{x \to 0} \frac{\sin^{2}{x}}{x} \\ &=-L \Rightarrow 2L =0 \Rightarrow L=0
\end{align*}
A: Using half-angle formula, we have $1-\cos x = 2\sin^2\frac x2$ so
$$
\lim\limits_{x\to 0}\frac{1-\cos x}{x} = 2\lim\limits_{x\to 0}\frac{\sin^2\frac x2}{x} = \lim\limits_{x\to 0}\frac{\sin\frac x2}{x/2}\cdot\lim\limits_{x\to 0}\sin\frac x2 = 1\cdot0 = 0
$$
where we used the fact that both limits exist.
A: Using L'Hopital 
$$\lim \limits_{x \to 0} \frac{1 - \cos x}{x} = \lim \limits_{x \to 0}\frac{(1 - \cos x)'}{x'} =  \lim \limits_{x \to 0} \sin x = 0$$
