The key here lies in the longterm behavior of $(2\sin x)^{2n}$. Note that $\sin x$ is fixed: the limit is over $n$, so what is changing is the exponent. So we are really looking at the behavior of $a^{2n}$ as $n\to\infty$, where $a$ is some number between $-2$ and $2$ (namely, $a=2\sin x$).
If $-1\lt 2\sin x\lt 1$, then $(\sin x)^{2n}\to 0$ as $n\to\infty$, so $f(x) = x$. On the other hand, if $2\sin x\gt 1$ or $2\sin x\lt -1$, then $(2\sin x)^{2n} = \left((2\sin x)^2\right)^n$ will go to $\infty$ as $n\to\infty$, so the limit will be $0$. Finally, if $2\sin x = \pm 1$, then $(2\sin x)^{2n}\to 1$ as $n\to\infty$, so the limit will be $1$. That is: $\lim_{n\to\infty}\frac{x}{(2\sin x)^{2n}+1} = \left\{\begin{array}{lll} x &&\text{if }|\sin x|\lt \frac{1}{2}\\ \frac{x}{2} &\quad&\text{if }|\sin x|=\frac{1}{2};\\ 0 && \text{if }|\sin x|\gt\frac{1}{2}. \end{array}\right.$ Is this enough for you to finish off the problem, analyzing the continuity of $f(x)$? If not, I'll add more.