Pointwise convergence - $\frac{nx}{1+n \sin(x)}$ , $x \in [0, \frac{\pi}{2}]$ Is anyone able to check if this is correct:
for $$f_n(x) = \frac{nx}{1+n \sin(x)} , x \in [0, \frac{\pi}{2}]$$
Does this converge pointwise to $$ \frac{x}{\sin(x)}$$
I am unsure due to the fact that at $x=0$ that would be undefined,
yet
$$\lim_{n\to \infty} \frac{nx}{1+n \sin(x)} = \lim_{n\to \infty} \frac{x}{\frac{1}{n}+ \sin(x)} = \frac{x}{\sin(x)}$$
I am not sure what else to do further, or how to handle this for the case of $x= 0$ ?
Can it be used the fact that $$\lim_{x \to 0} \frac{x}{\sin(x)} = 1$$
Any help is appreciated.

You have already proved that
$$\lim_{n\to\infty} f_n(x) = \begin{cases}0 & x=0\\ \frac x{\sin x} & x \ne 0 \end{cases}$$
And this limit is indeed pointwise (as you can fix $x\ne 0$ and obtain the limit $\frac x{\sin x}$ and fix $x=0$ and obtain $f_n(0) = 0 \to 0$).
Furthermore because the $f_n$ are continuous for $x\in [0, \frac\pi2]$ and the limit function is discontinuous at $0$ (as you correctly asserted), you can also say that the convergence is not uniform.
