why is it the lim?

• Mar 29th 2013, 04:42 AM
orir
why is it the lim?
$f(x)=(sinx)^2/ (x^3-pi*x^2)$

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, 05:44 AM
Ruun
Re: why is it the lim?
For

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

You can use L'Hôpital's rule

$\lim_{x\to \pi}\frac{2\sin(x)\cos(x)}{2x(x-\pi)+x^2}$
• Mar 29th 2013, 05:53 AM
orir
Re: why is it the lim?
Quote:

Originally Posted by Ruun
For

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

You can use L'Hôpital's rule

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

i didn't learn this rule yet, and i suppose i'm not allowed to use it before i do.
is there another (may be complicated, but regular) way to solve this?
• Mar 29th 2013, 05:56 AM
Ruun
Re: why is it the lim?
Do you know how to solve $0\cdot\infty$ limits?
• Mar 29th 2013, 06:13 AM
orir
Re: 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, 06:39 AM
emakarov
Re: why is it the lim?
\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, 07:08 AM
orir
Re: why is it the lim?
thank you so much!