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How to find $$\lim_{x\rightarrow 0} \frac{\sin |x|}{x^2+\sin (x)}$$ without L'hopital's?

So far I tried to use the squeeze theorem but couldn't find appropriate bounds and also tried to exploit the limit of $\sin(x)/x$ without any luck.

Any hints?

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  • $\begingroup$ Question is: does it have a limit? $\endgroup$ – user228113 Apr 22 '16 at 15:27
  • $\begingroup$ I'm not sure to be honest. $\endgroup$ – Chris Apr 22 '16 at 15:28
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    $\begingroup$ "and also tried to exploit the limit of $\sin(x)/x$ without any luck" This lack of "luck" is odd since you are looking at $$\frac{\sin |x|}{|x|}\frac{1}{x+\frac{\sin (x)}{x}}\frac{x}{|x|},$$ whose behaviour when $x\to0$ is pretty clear. $\endgroup$ – Did Apr 22 '16 at 15:29
  • $\begingroup$ @G.Sassatelli Let me suggest you compare $$\frac{|x|}{x}\quad\text{and}\quad\frac{x}{|x|}...$$ Or is your comment tongue-in-cheek? $\endgroup$ – Did Apr 22 '16 at 15:32
  • $\begingroup$ @Did Oh! Silly me. It was just me looking at symbols instead of evaluating the function. $\endgroup$ – user228113 Apr 22 '16 at 15:35
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You can write the function as $$ \frac{\sin|x|}{x}\frac{1}{x+\dfrac{\sin x}{x}} $$ The second factor has limit $1$ for $x\to0$, so the problem is reduced to seeing whether $$ \lim_{x\to0}\frac{\sin|x|}{x} $$ exists.

Hint: try from the left and from the right.

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  • $\begingroup$ Limit doesn't exist since left hand limit is $-1$ whereas right hand limit is $1$? $\endgroup$ – Chris Apr 22 '16 at 15:39
  • $\begingroup$ @Chris Exactly so. $\endgroup$ – egreg Apr 22 '16 at 15:41
  • $\begingroup$ @Chris Note that applying l'Hôpital would give a non existing limit either: the function would be $\dfrac{|x|\cos|x|}{x(2x+\cos x)}$, but this would not be immediately conclusive. On the other hand, this would give limit $-1$ from the left and $1$ from the right, so the same as with the direct method. $\endgroup$ – egreg Apr 22 '16 at 15:44
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For another approach, look at the reciprocal:

Set

$h(x)=\frac{x^2+\sin x}{\sin \vert x\vert}$.

Then,

$h(x) =x\frac{x}{\sin \vert x\vert}+\frac{\sin x}{\sin \vert x\vert }$.

and

$h(x)\to 0+1=1$ as $x\to 0^+$

whereas

$h(x)\to 0-1=-1$ as $x\to 0^-$

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