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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,

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