We first need to determine lower/upper functions. We’ll start off by acknowledging that provided $x \ne 0$ (which we know it won’t be because we are looking at the limit as$x \to 0$) we will have,

$$- 1 \le \sin \left( {\frac{\pi }{x}} \right) \le 1$$

Now, simply multiply through this by ${x^4}$ to get,

$$- {x^4} \le {x^4}\sin \left( {\frac{\pi }{x}} \right) \le {x^4}$$

Before proceeding note that we can only do this because we know that ${x^4} > 0$ for $x \ne 0$. Recall that if we multiply through an inequality by a negative number we would have had to switch the signs. So, for instance, had we multiplied through by ${x^3}$ we would have had issues because this is positive if $x > 0$ and negative if $x < 0$.

Now, let’s get back to the problem. We have a set of lower/upper functions and clearly,

$$\mathop {\lim }\limits_{x \to 0} {x^4} = \mathop {\lim }\limits_{x \to 0} \left( { - {x^4}} \right) = 0$$

Therefore, by the Squeeze Theorem we must have,

$$\mathop {\lim }\limits_{x \to 0} {x^4}\sin \left( {\frac{\pi }{x}} \right) = 0$$