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$$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$

now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}=\left({\sin\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)}\right)^2=1^2$$

So we have $$\frac{2}{x^2}\cdot \left(\frac{x}{2}\right)^2=\frac{2}{x^2}\cdot \left(\frac{x^2}{4}\right)=\frac{1}{2}$$

Are the moves right?

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    $\begingroup$ You'd be correct if you wrote $\lim_{x\to 0}$ on every expression. $\endgroup$ – user26486 Jun 21 '15 at 20:16
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    $\begingroup$ Another way you could do this is to multiply by $1+\cos x$ on the top and bottom. $\endgroup$ – user84413 Jun 21 '15 at 20:17
  • $\begingroup$ Taylor expansion is probably the easiest way. It's a definition and can easily yield 1/2. Also shows $x^3$ in the denominator can't work. $\endgroup$ – user228288 Jun 21 '15 at 20:30
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Correct, but too complicated (and missing several $\lim_{x\to0}$). $$ \lim_{x\to0}\frac{1-\cos x}{x^2}= \lim_{x\to0}\frac{2\sin^2(x/2)}{4(x/2)^2}= \lim_{x\to0}\frac{1}{2}\left(\frac{\sin(x/2)}{(x/2)}\right)^{\!2}= \frac{1}{2} $$

Alternative way: $$ \lim_{x\to0}\frac{1-\cos x}{x^2}= \lim_{x\to0}\frac{1-\cos^2 x}{x^2(1+\cos x)}= \lim_{x\to0}\frac{1}{1+\cos x}\left(\frac{\sin x}{x}\right)^{\!2} $$

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  • $\begingroup$ Thanks for mentioning the alternative way as it is not very well known. It is much better because it avoid the formula connecting $\sin x$ with $\sin (x/2)$ and is at a much simpler level. +1 $\endgroup$ – Paramanand Singh Jun 22 '15 at 5:44
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I'm surprised nobody used L'Hopital's Rule:

$$\lim _{x \to 0}{1-\cos x\over x^2}\equiv \lim _{x \to 0}{\sin x\over 2x}\equiv\lim _{x \to 0}{\cos x\over 2}=\frac{1}{2} $$

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To provide a correction to your own work I would remove the $\lim$ at first because I want to simplifies to the maximum the expression and at the last the computation, as follows: $${1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2 ={\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}\cdot \frac{1}{2}$$ therefore $$\lim{1-\cos x\over x^2}=\lim{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}\cdot \frac{1}{2}=1\cdot\frac{1}{2}=\frac{1}{2}.$$

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