Answer
$$\lim_{x\to0}h(x)=0$$
Work Step by Step
$$h(x)=x^2\cos\Big(\frac{1}{x^3}\Big)$$
(a) The graph is shown below.
We see from the graph that as $x\to0$, the value of $h(x)$, though still oscillates, gets closer and closer to a single value $y$ of $0$.
This means we can estimate here that $\lim_{x\to0}h(x)=0$
(b) To prove our estimation, we cannot use normal methods or manipulate limit laws. We need to take advantage of the Sandwich Theorem.
We know that $-1\le\cos\Big(\frac{1}{x^3}\Big)\le1$
So, $-x^2\le x^2\cos\Big(\frac{1}{x^3}\Big)\le x^2$ ($x^2\ge0$ for all $x$)
In other words, $-x^2\le h(x)\le x^2$
We have $\lim_{x\to0}x^2=\lim_{x\to0}(-x^2)=0$
Therefore, according to Sandwich Theorem, $\lim_{x\to0}h(x)=0$
