Let's start from $\dfrac{k^2\sin\left(\tfrac{1}{k}\right)}{k}=k \cdot \sin\left(\tfrac{1}{k}\right)$
At this point we need to look at a graph, SEE HERE
Looking at that graph, what does the limit as $k\to 0$ look like?
How do you prove it? If you need hints here they are:
1) $-1\le\sin\left(\tfrac{1}{k}\right)\le 1$
2) If $k>0$ then $-k\le k\sin\left(\tfrac{1}{k}\right)\le k$ and we know $k\to 0^+$
This is the tricky one, what about $k\to 0^-~?$
3) Well $\sin$ is an odd function so $\sin\left(\tfrac{1}{-k}\right)=-\sin\left(\tfrac{1}{|k|}\right)$ as can be seen on the graph.
In other words: the limit as $k\to 0^+$ is equal to the limit as $k\to 0^-$
To be fair to myself, I did say that this was tricky.
Odd functions are just that.
$\sin(-t)=-\sin(t)$ or $\sin(t)=-\sin(-t)$
THUS $\displaystyle \large{\mathop {\lim }\limits_{t \to {0^ - }} \sin (t) = \mathop {\lim }\limits_{t \to {0^ - }} - \sin ( - t) = \mathop {\lim }\limits_{t \to {0^ + }} - \sin (t)}$