Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 generalized it?

share|cite|improve this question
HINT: $1-\cos(x) = 2 \sin^2\left(\frac{x}{2}\right)$ – Sasha Dec 1 '11 at 20:23
With this particular trick, yes: use the half-angle formula. For another trick, multiply and divide by a clever $1$: $$1 = \frac{1+\cos(ax)}{1+\cos(ax)}.$$Or make a change of variable, $u=ax$. – Arturo Magidin Dec 1 '11 at 20:23
Do you know how to prove $\lim \limits_{x \to 0}\frac{1-\cos(x)}{x}=0$ ? If yes, then as $x \to 0 \Rightarrow ax \to 0$ ... – Quixotic Dec 1 '11 at 20:25
@Pedro: $\LaTeX$ tips: rather than using \being{align*}...\end{align*} in an single-line in-line formula, you can use \displaystyle to get it to show in display format. You can also use \limits: \lim\limits_{x\to 0} produces $\lim\limits_{x\to 0}$, even in in-line formulas. Also, use \sin and \cos for the trig functions. – Arturo Magidin Dec 1 '11 at 20:32

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 $$ \frac{1-\cos t}{t} = \frac{o(t)}{t} = 0. $$

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.