How to find $\lim_{x\rightarrow 0} \frac{\sin |x|}{x^2+\sin (x)}$?

Question

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?

Answer 1

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.

Answer 2

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 $x\to 0^-$