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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?

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HINT: $1-\cos(x) = 2 \sin^2\left(\frac{x}{2}\right)$ –  Sasha Dec 1 '11 at 20:23
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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
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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
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@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

2 Answers 2

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

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

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