Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 63

Prove that the function
$f(x) = \left\{ \begin{array}{cc} x^4 \sin \left( \frac{1}{x} \right) & \text{ if } x \neq 0 \\ 0 & \text{ if } x = 0 \end{array} \right.$ is continuous everywhere.

Using the Squeeze Theorem to prove that the left and right hand limits are equal...


$\begin{equation} \begin{aligned} & -1 \leq \sin \left( \frac{1}{x} \right) \leq 1\\ \\ & -x^4 \leq x^4 \sin \left( \frac{1}{x} \right)\leq x^4\\ \\ & \lim \limits_{x \to 0} (-x^4) = -(0)^4 = 0\\ \\ & \lim \limits_{x \to 0} (x^4) = 0^4 = 0 \end{aligned} \end{equation}$


The Squeeze Theorem gives us $\displaystyle\lim \limits_{x \to 0} x^4 \sin \left( \frac{1}{x} \right) = f(0) = 0$. Therefore, the given function is continuous on $(-\infty, \infty)$