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Math Help - limit?

  1. #1
    Super Member Showcase_22's Avatar
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    limit?

    Find the limit (if it exists) of \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}.
    It's apparent that \sin \left( \frac{1}{x} \right) is discontinuous (I have a proof for that involving the sequence a_n=\frac{1}{2n \pi+\frac{\pi}{2}} and b_n=\frac{1}{2n \pi -\frac{\pi}{2}}).

    Define -\lim_{x \rightarrow 0} \frac{x^2}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2}{\sin x}

    \lim_{x \rightarrow 0} \frac{x^2}{\sin x}=\lim_{x \rightarrow} \frac{2x}{\cos x}=0

    Hence \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}=0

    Is this the correct way of doing this type of question or is there an easier way?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Showcase_22 View Post
    It's apparent that \sin \left( \frac{1}{x} \right) is discontinuous (I have a proof for that involving the sequence a_n=\frac{1}{2n \pi+\frac{\pi}{2}} and b_n=\frac{1}{2n \pi -\frac{\pi}{2}}).
    Trying this as \left(\frac{x}{sin x}\right)\left(\frac{sin(1/x)}{(1/x)}\right) should make it much easier! As x goes to 0, 1/x goes to infinity while sin(1/x) is always between -1 and 1.

    Define -\lim_{x \rightarrow 0} \frac{x^2}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x} \leq \lim_{x \rightarrow 0} \frac{x^2}{\sin x}

    \lim_{x \rightarrow 0} \frac{x^2}{\sin x}=\lim_{x \rightarrow} \frac{2x}{\cos x}=0

    Hence \lim_{x \rightarrow 0} \frac{x^2 \sin \left( \frac{1}{x} \right)}{ \sin x}=0

    Is this the correct way of doing this type of question or is there an easier way?
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