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…
Best Answer
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.