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Section 2.5 : Computing Limits

15. Use the Squeeze Theorem to determine the value of \(\mathop {\lim }\limits_{x \to 0} {x^4}\sin \left( {\frac{\pi }{x}} \right)\).

Hint : Recall how we worked the Squeeze Theorem problem in this section to find the lower and upper functions we need in order to use the Squeeze Theorem.
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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,

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