Continuity check

If $x\ne0$ , $f(x)$ is a geometric series whose first term is $x^m$ , and common ratio $\dfrac{1}{1+x^4}<1$ . Therefore, $f(x)=\dfrac{x^m}{1-\dfrac{1}{1+x^4}}=\dfrac{x^m(1+x^4)}{x^4}=x^m(1+1/x^4)$ .

If $x=0$ , $f(0)=0$ .

$\therefore f(x)=\left\{\begin{array}{ll}x^m(1+1/x^4) & \mathrm{if}\; x\ne 0 \\ 0& \mathrm{if}\; x=0\end{array}\right.$

Since $f$ is continuous at $x=0$ , $\displaystyle \lim_{x\to0}x^m(1+1/x^4)=0$ . If $1\le m \le 3$ , the limit doesn't exist. If $m=4$ , the limit is 1. If $m\ge 5$ , the limit is 0 and we get the answer.