$\displaystyle f(x)=(sinx)^2/ (x^3-pi*x^2)$

why is lim x->pi f(x)=0?

(does it have something to do with the fact that lim x->0 of sinx/x equals 1?)

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- Mar 29th 2013, 03:42 AMorirwhy is it the lim?
$\displaystyle f(x)=(sinx)^2/ (x^3-pi*x^2)$

why is lim x->pi f(x)=0?

(does it have something to do with the fact that lim x->0 of sinx/x equals 1?) - Mar 29th 2013, 04:44 AMRuunRe: why is it the lim?
For

$\displaystyle \lim_{x\to \pi}\frac{\sin^2(x)}{x^2(x-\pi)}$

You can use L'Hôpital's rule

$\displaystyle \lim_{x\to \pi}\frac{2\sin(x)\cos(x)}{2x(x-\pi)+x^2}$ - Mar 29th 2013, 04:53 AMorirRe: why is it the lim?
- Mar 29th 2013, 04:56 AMRuunRe: why is it the lim?
Do you know how to solve $\displaystyle 0\cdot\infty$ limits?

- Mar 29th 2013, 05:13 AMorirRe: why is it the lim?
no, i don't.. (Thinking)

(i acctually thought it can't be done and we must change the function's form to solve this... so it can be done?) - Mar 29th 2013, 05:39 AMemakarovRe: why is it the lim?
$\displaystyle \begin{align*}\lim_{x\to\pi}\frac{\sin^2x}{x^3-\pi x^2} &= \lim_{x\to\pi}\left(\frac{\sin x}{x-\pi}\cdot\frac{\sin x}{x^2}\right)\\ &= \left(\lim_{x\to\pi}\frac{\sin x}{x-\pi}\right)\left(\lim_{x\to\pi}\frac{\sin x}{x^2}\right)\\ &= \left(\lim_{x\to\pi}\frac{-\sin (x-\pi)}{x-\pi}\right)\cdot0\\ &= \left(\lim_{y\to0}-\frac{\sin y}{y}\right)\cdot0\\ &= -1\cdot0=0\end{align*}$

- Mar 29th 2013, 06:08 AMorirRe: why is it the lim?
thank you so much!