evaluate the Limit $$ \lim \limits_{z \to 0} z\sin(1/z^2)$$
Anyone can help me with this question? Not sure how to solve this. I tried to bring z to denominator but don know how to continue.
 A: Note that $\sin$ is a bounded function, so:
$$
\lvert z\sin(1/z^2) \rvert \leq \lvert z\rvert
$$ So when you multiply something that is bounded by something that approaches zero, then the whole goes to ... 
A: Given $$\displaystyle \lim_{x\rightarrow 0}z\cdot \sin \left(\frac{1}{z^2}\right)$$
Now we know that $$\displaystyle - 1\leq \sin \left(\frac{1}{z^2}\right)\leq 1$$
So $$\displaystyle  \lim_{z\rightarrow 0}-z\leq \lim_{x\rightarrow 0}z\cdot \sin \left(\frac{1}{z^2}\right)\leq \lim_{z\rightarrow 0}z$$
So here $\bf{Left}$ side inequality and $\bf{Right}$ side inequality is $=0\;,$ When $z\rightarrow 0$
So Using Sandwitch Theorem, We get $$\displaystyle \lim_{z\rightarrow 0}z\cdot \sin \left(\frac{1}{z^2}\right) =0$$
A: A slight variation:  Let y= 1/z.  As z goes to 0, y goes to infinity so this limit becomes $\lim_{y\to\infty} \frac{sin(y^2)}{y}$.  Now, as juantheron and Thomas said, sin(y) is always between -1 and 1:
$$-\frac{1}{y}< \frac{sin(y^2)}{y}< \frac{1}{y}$$
and both ends go to 0 as y goes to infinity.
