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...

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.

$$ \begin{eqnarray*} \lim_{x \to 0} \frac{1 - \cos{x}}{x} &=& \lim_{x \to 0} \frac{(1-\cos{x})(1+\cos x)}{x(1+\cos x)} \\ &=& \lim_{x \to 0} \frac{1-\cos^2 x}{x(1 + \cos x)} \\ &=& \lim_{x \to 0} \frac{x\sin^2 x}{x \cdot x(1+ \cos x)} \\ &=& \lim_{x \to 0} \frac{\sin x}{x} \times \frac{\sin{x}}{x}\times \frac{x}{1+\cos x} = 0 . \end{eqnarray*} $$