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 $(-\infty, \infty)$