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I want to show that the function $\displaystyle f(x)= \frac{1-\cos x}{x^{2}}$ is bounded in $(-\infty,\infty)$. I know that $\displaystyle h(x)=1-\cos x$ is bounded on $R$ but $\displaystyle g(x)=\frac{1}{x^{2}}$ is not bounded in nbhd. of $0$. So what about $h(x).g(x)$ on $R$? Is it bounded? Why?

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    Clearly the problem region is a neighborhood about $x=0$, and this problem is really just a question of order of magnitudes, e.g. which part of the quotient is decaying to $0$ faster as $x\to0$. Loosely speaking, if the denominator decays faster, then it is not bounded; if the numerator decays faster, then it is bounded; if they decay a relatively comparable rates, then it is still bounded by some absolute constant $M$. Now you could estimate the orders yourself, but this problem screams L'Hopitals Rule as alluded to by Andre.2012-10-28

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Hint: Find $\lim_{x\to 0}\frac{1-\cos x}{x^2}.$

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    As to avoiding using limits, it is clear that if we take any explicit interval $(-a,a)$, for any positive $a$, then our function is bounded outside $(a,a)$. So the only issue is the behaviour very near $0$. One can find inequalities that deal with the matter, using the double angle identity $1-\cos x=2\sin^2(x/2)$, and the fact that $|\sint t|\le |t|$ to get a bound. That is awfully close to what we do when we find the limit.2012-10-28
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$\begin{eqnarray*} \frac{1-\cos x}{x^2} &=& \frac{2\sin^2\frac{x}{2}}{x^2} \end{eqnarray*}$ Now note that $\sin^2 x \le x^2$.