TITLE: evaluate the Limit
QUESTION [2 upvotes]: $$ \lim \limits_{z \to 0} z\sin(1/z^2)$$
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.

REPLY [0 votes]: 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.
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