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Math Help - why is it the lim?

  1. #1
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    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?)
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  2. #2
    Member Ruun's Avatar
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    Re: why is it the lim?

    For

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

    You can use L'H˘pital's rule

    \lim_{x\to \pi}\frac{2\sin(x)\cos(x)}{2x(x-\pi)+x^2}
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  3. #3
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    Re: why is it the lim?

    Quote Originally Posted by Ruun View Post
    For

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

    You can use L'H˘pital'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?
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  4. #4
    Member Ruun's Avatar
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    Re: why is it the lim?

    Do you know how to solve 0\cdot\infty limits?
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  5. #5
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    Re: why is it the lim?

    no, i don't..
    (i acctually thought it can't be done and we must change the function's form to solve this... so it can be done?)
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  6. #6
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    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*}
    Thanks from Ruun
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  7. #7
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    Re: why is it the lim?

    thank you so much!
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