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It is a well know result that $$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$$ so we should have the result that $$\lim_{x\rightarrow 0} x^2\sin \left(\frac{1}{x^2}\right)=1$$ yet when I plug the limit into an online calculator it uses the squeeze theorem and shows that $$-x^2 \leq x^2\sin \left(\frac{1}{x^2}\right) \leq x^2 $$ because of the values $\sin$ can take. This leads us to the result that the limit is zero because of the limits of $x^2,-x^2$ as $x$ tends to zero.

Why is the middle line wrong though? I've been sat for a while thinking about it but I can't see why I'm guessing there is something obvious I'm not spotting.

Thanks.

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    $\begingroup$ Why do you think that $\lim_{x\rightarrow 0} x^2\sin \left(\frac{1}{x^2}\right)=1$? $\endgroup$ – Aditya Agarwal Oct 14 '15 at 15:50
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    $\begingroup$ $x \to 0$ in the first limit corresponds to $x \to \infty$ in the second. $\endgroup$ – Robert Israel Oct 14 '15 at 15:50
  • $\begingroup$ Ah okay I get it now. Thanks. $\endgroup$ – Ryan Oct 14 '15 at 15:53
  • $\begingroup$ Robert please post yours as an answer and I will approve. $\endgroup$ – Ryan Oct 14 '15 at 15:55
  • $\begingroup$ @RobertIsrael That should really be an answer, not a comment... $\endgroup$ – 5xum Oct 14 '15 at 16:11
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As requested, I'm making my comment into an answer:

$x \to 0$ in the first limit corresponds to $x \to \infty$ in the second.

I should make that a bit more precise. It is better to use a different name for the variable in the other limit. With $t = 1/\sqrt{|x|}$ $$\eqalign{\lim_{x \to 0+} \dfrac{\sin(x)}{x} &= \lim_{t \to \infty} t^2 \sin(1/t^2)\cr \lim_{x \to 0-} \dfrac{\sin(x)}{x} &= \lim_{t \to \infty} -t^2 \sin(-1/t^2) = \lim_{t \to \infty} t^2 \sin(1/t^2) \cr}$$

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Robert has the correct answer in the comments: $x\to 0$ in the first limit corresponds to $x\to \infty$ in the second limit, so the analogy is incorrect.

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$$\lim_{x\rightarrow 0} x^2\sin \left(\frac{1}{x^2}\right)=0$$ because $\left|\sin \left( \frac{1}{x^2} \right) \right|\leq 1$ and $\lim\limits_{x\rightarrow 0} x^2=0$

Regarding $\lim\limits_{x\rightarrow 0} \frac{\sin(x)}{x}=1$ For one that knows the L'Hospital Rule it is simple $$\lim\limits_{x\rightarrow 0} \frac{\sin(x)}{x}= \lim\limits_{x\rightarrow 0} \frac{\cos(x)}{1}=1$$

Without the L'Hospital Rule go to Link it user some geometry basics and squeeze theorem, shouldn't not that difficult to understand. Another option is Another Link.

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  • $\begingroup$ This answer does not really address OP's main source of confusion, though... $\endgroup$ – 5xum Oct 14 '15 at 16:12
  • $\begingroup$ @Michael Medvinsky. It is not appropriate to use L'Hospital's Rule for this limit. Your link is the right way $\endgroup$ – imranfat Oct 14 '15 at 16:56
  • $\begingroup$ @imranfat this why the link exists $\endgroup$ – Michael Medvinsky Oct 14 '15 at 17:42
  • $\begingroup$ @5xum didn't get you $\endgroup$ – Michael Medvinsky Oct 14 '15 at 17:42

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