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Math Help - Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0

  1. #1
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    Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0

    Hello MHF, please help me how to proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0, the book says that L'Hospital Rule doenst work in this case.

    Thanks!
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  2. #2
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    Re: Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0

    Quote Originally Posted by Chipset3600 View Post
    Hello MHF, please help me how to proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0, the book says that L'Hospital Rule doenst work in this case.

    Thanks!
    Note that -1 \le \sin(\left( \frac{1}{x}\right) \le 1


    Now if we multiply by \frac{x^2}{\sin(x)}  we get

    -\frac{x^2}{\sin(x)} \le \frac{x^2\sin\left( \frac{1}{x}\right)}{\sin(x)} \le \frac{x^2}{\sin(x)}

    Note that the limit

    \lim_{x \to 0}\frac{x^2}{\sin(x)}=0

    So by the squeeze theorem the limit is 0
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0

    I would suggest splitting the fraction in two - we know that the limit of x/sin(x) as x goes to 0 is 1. So that leaves the limit of xsin(1/x). The value of sin(1/x) fluctuates between -1 and 1 as x goes to 0. So the limit of x times that is 0. Hence the answer is 0.
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  4. #4
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    Re: Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0

    wow! understood, I would not find this solution so quickly. Thanks guys
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