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So this is the question in my text book $$\lim_{x\to0}\frac{x^2\sin\left(\frac{1}{x}\right)}{\sin x}$$

here what i have done

It can be rearranged as $$\lim_{x\to0}\left(\frac{x}{\sin x}\right)x\sin\left(\frac{1}{x}\right)$$$$\Rightarrow\lim_{x\to0}(1)x\sin\left(\frac{1}{x}\right)$$$$\Rightarrow\lim_{x\to0}\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$$$$\Rightarrow1$$

This is what i got. Have done something wrong cause in my text-book the answer is $0$

Here the solution in my text book $$\lim_{x\to0}\left(\frac{x}{\sin x}\right)x\sin\left(\frac{1}{x}\right)$$$$\Rightarrow\lim_{x\to0}(1)x\sin\left(\frac{1}{x}\right)$$$$\Rightarrow0$$ as$$\left|x\sin\frac{1}{x}\right|\le|x|$$

can anyone help me what is wrong

Thanks

Akash

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    $\begingroup$ It is not true that $\lim_{x\to 0^+}\frac{\sin(1/x)}{1/x}=1$. For $1/x$ does not approach $0$. The important thing is that $\sin(1/x)$ is bounded, so $x\sin(1/x)\to 0$ as $x\to 0^+$. $\endgroup$ – André Nicolas Jan 12 '14 at 4:03
  • $\begingroup$ thanks for the explanation nice $\endgroup$ – Dimensionless Jan 12 '14 at 4:29
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    $\begingroup$ You are welcome. That is something that comes up medium often with questions that involve sine or cosine. $\endgroup$ – André Nicolas Jan 12 '14 at 5:13
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In your approach, you claim that $\lim_{x \rightarrow 0}\dfrac{\sin\frac{1}{x}}{\frac{1}{x}}=1$, but in fact the limit $\dfrac{\sin t}{t} \rightarrow 1$ just holds for $t \rightarrow 0$. In your solution $\dfrac{1}{x}$ doesn't satisfy the condition $\rightarrow 0$ but $\rightarrow \infty$.

A correct solution may be the following:

Since $\sin\dfrac{1}{x}$ is bounded and $x \rightarrow 0$, $\lim_{x \rightarrow 0}x \sin\frac{1}{x}=0$.

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  • $\begingroup$ @Akash You're welcome! $\endgroup$ – Xucheng Zhang Jan 12 '14 at 4:34
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$$\lim_{x\to0}\frac{x^2\sin\frac1x}{\sin x}=\lim_{x\to0}\frac{x\sin\frac1x}{\frac{\sin x}x}=\lim_{x\to0}\frac{x\sin\frac1x}1=0.$$

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  • $\begingroup$ Which is so because $\sin \frac 1x$ is bounded and $x\sin \frac 1x\to 0$ as $x\to0$ $\endgroup$ – abiessu Jan 12 '14 at 4:04
  • $\begingroup$ @Akash: $$-1\le\sin\frac1x\le1\iff-x\le x\sin\frac1x\le x\iff0\le\lim_{x\to0}x\sin\frac1x\le0\iff\lim_{x\to0}x\sin\frac1x=0$$ $\endgroup$ – Lucian Jan 12 '14 at 4:41
  • $\begingroup$ The first rightward implication is true for $\;x>0\;$ . Not that it really matters in the final outcome as you can work through the squeeze theorem once with $\;x>0\;$ and once with $\;x<0\;$ ... $\endgroup$ – DonAntonio Jan 12 '14 at 5:20

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