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,

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