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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 generalize it?

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    __HINT__: $1-\cos(x) = 2 \sin^2\left(\frac{x}{2}\right)$2011-12-01
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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, $u=ax$.2011-12-01
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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 $x \to 0 \Rightarrow ax \to 0$ ...2011-12-01
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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.2011-12-01

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