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

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 AM #1

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- Mar 29th 2013, 04:44 AM #2

- Mar 29th 2013, 04:53 AM #3

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- Mar 29th 2013, 04:56 AM #4

- Mar 29th 2013, 05:13 AM #5

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- Mar 29th 2013, 05:39 AM #6

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## Re: 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 AM #7

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