Answer
The function is continuous on $(-\infty, \infty)$.
Work Step by Step
$f(x) = x^4~sin(\frac{1}{x})$, if $x \neq 0$
Clearly this function is continuous on the intervals $(-\infty, 0)$ and $(0, \infty)$
Since $f(0) = 0$, we need to show that $\lim\limits_{x \to 0}x^4~sin(\frac{1}{x}) = 0$
Let $\epsilon \gt 0$ be given.
Let $\delta = \epsilon^{1/4}$
Suppose that $\vert x - 0 \vert \lt \delta$ (but $x \neq 0$)
Then:
$\vert x^4~sin(\frac{1}{x}) \vert = \vert x^4 \vert \cdot \vert sin(\frac{1}{x})\vert \lt \delta^4 \cdot 1 = \epsilon$
Therefore, $\lim\limits_{x \to 0}x^4~sin(\frac{1}{x}) = 0$
Thus, the function is continuous on $(-\infty, \infty)$
