# Trigonometric limit: $\lim_{x \to 0}\frac{1-\cos(ax)}{ax}=0$

In order to prove that $\displaystyle\lim_{x \to 0}\frac{1-\cos(ax)}{ax}=0$, with $a \ne 0$, I managed that $a=2$ and evaluated this limit:

\begin{align*} \quad \lim_{x \to 0}\frac{1-\cos(2x)}{2x}&= \lim_{x \to 0}\frac{1-(1-2\sin^2(x))}{2x}\\ &= \lim_{x \to 0}\frac{1-1+2\sin^2(x)}{2x}\\ &= \lim_{x \to 0}\frac{2\sin^2(x)}{2x}\\ &= \lim_{x \to 0}\frac{\sin^2(x)}{x}\\ &= \lim_{x \to 0} \frac{\sin(x)}{x} \cdot \sin(x)\\ &= 1 \cdot 0\\ &=0 \end{align*}

Can I generalize it?

#### Solutions Collecting From Web of "Trigonometric limit: $\lim_{x \to 0}\frac{1-\cos(ax)}{ax}=0$"

As written in comments you can use the fact that $x \to 0$ if and only if $ax \to 0$ and make a substitution $t = ax$. Then your limit takes form
$$\lim_{t \to 0} \frac{1-\cos t}{t}.$$
Next, using Taylor expansion $\cos t = 1 – t^2/2 + t^4/4! – t^6/6! + \dots = 1 + o(t)$ you get
$$\frac{1-\cos t}{t} = \frac{o(t)}{t} = 0.$$

Just apply L’Hopital Rule. More concretely,
$$\lim_{x\rightarrow 0} \frac{1- \cos ax}{ax} = \lim_{x \rightarrow 0} \frac{a\sin ax}{a} =0.$$