Testing continuity of the function $f(x) = \lim\limits_{n \to \infty} \frac{x}{(2\sin{x})^{2n}+1} \ \text{for} \ x \in \mathbb{R}$

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.

Proceeding from where Arturo left note that:

• $|2\sin{x}|< 1$ $\Longrightarrow$ $x \in \displaystyle\Bigl(n\pi-\frac{\pi}{6},n\pi+\frac{\pi}{6}\:\Bigr)$

• $|2\sin{x}|=1 \Longrightarrow \displaystyle x =n\pi+\frac{\pi}{6}$

• $|2\sin{x}|>1 \Longrightarrow \displaystyle x \in \Bigl(n\pi + \frac{\pi}{6}, n\pi+\frac{5\pi}{6}\:\Bigr)$

So your $f(x)$ is basically, $$ f(x) = \left\{\begin{array}{lll} x & \text{if}\ x\in \Bigl(n\pi-\frac{\pi}{6},n\pi+\frac{\pi}{6}\Bigr) \\\ \frac{x}{2} & \text{if} \ x =n\pi+\frac{\pi}{6} \\\ 0 & \text{if} \ x \in \Bigl(n\pi + \frac{\pi}{6}, n\pi+\frac{5\pi}{6}\Bigr)\end{array}\right.$$

It is easy to note $\text{that the only points which gives us trouble are}$ $x =n\pi \pm \frac{\pi}{6}$. Now it is your job is to check the continuity of $f(x)$ at $x=n\pi \pm \frac{\pi}{6}$.