Answer
$f$ is continuous on $(-\infty,\infty)$.
Work Step by Step
\[f(x)=\left\{\begin{array}{cc}x^4\sin\left(\frac{1}{x}\right)\;\;\;\; x\neq 0\\
0\;\;\;\;\;\;\;\;\;\; x=0\end{array}\right.\]
Clearly $f$ is continuous if $x\neq 0$ so we will check the continuity at $x=0$.
Let,
\[L=\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0^{-}}x^4\sin\left(\frac{1}{x}\right)\]
\[-x^4\leq x^4\sin\left(\frac{1}{x}\right)\leq x^4\]
Also $\displaystyle\lim_{x\rightarrow 0}(-x^4)=0=\lim_{x\rightarrow 0}x^4$
By sandwich theorem,
\[L=0\]
\[\Rightarrow \lim_{x\rightarrow 0}f(x)=f(0)=0\]
$\Rightarrow f(x)$ is continuous at $x=0$
Therefore $f$ is continuous on $(-\infty,\infty)$.
Hence Proved.
