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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. – Taylor Martin Oct 28 '12 at 4:35
up vote 4 down vote accepted

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

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Its limit is $\frac{1}{2}$. I know that the limit of function exist then it is bounded. But if i want to show that this function is bounded without using the concept of limit then how can i show? – Kns Oct 28 '12 at 4:34
Why would you want to avoid using limits? There are lots of alternative (though far less direct ways to do this): (1) The function is continuous on any compact subset of $\mathbb{R}$ (even at the origin); prove this and use the maximal principle of continuous functions on compact sets combined with the obvious fact that outside of some interval $[-M,M]$ the function is bounded. (2) The function is differentiable (prove this); then compute its maximal value and show that it is less than $\infty.$ (3) Compare $f$ to a function $g\geq f$ which is known to be bounded about the origin. – Taylor Martin Oct 28 '12 at 4:41
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. – André Nicolas Oct 28 '12 at 5:32

$$\begin{eqnarray*} \frac{1-\cos x}{x^2} &=& \frac{2\sin^2\frac{x}{2}}{x^2} \end{eqnarray*}$$ Now note that $\sin^2 x \le x^2$.

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