TITLE: Proof check, showing pointwise convergence
QUESTION [5 upvotes]: My problem is this:
For $x \in [0,\frac{\pi}{2}]$, $f_n(x) = \frac{nx}{1 + n\sin(x)}$
Find the pointwise limit of $(f_n)$ for all $x \in [0, \frac{\pi}{2}]$

I am not sure if the way I constructed my "proof" is correct, but split it into cases, as follows:
Case 1: $x \in (0, \frac{\pi}{2}]$
$$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{nx}{n \sin(x) + 1} = \frac{x}{\sin(x)}$$
Case 2: $x=0$
$$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{n(0)}{1 + 0} = 0$$
Therefore, 
$$\lim_{n \to \infty} f_n(x) = f(x)$$
where $$f(x) = 0 , x= 0$$ and $$ f(x) = \frac{x}{\sin(x)}, x \in (0, \frac{\pi}{2}]$$
Is there any way to improve how I have constructed this ?
I was a little unsure with the choice of doing a "case 1/case 2" for this question.
Any tips are appreciated

REPLY [2 votes]: Your proof is fine, but I added some points and shortened it.
You can say that $f_n \rightarrow f$ pointwise, where 
$f = \begin{cases} 0 & \text{if } x = 0 \\ \frac{x}{\sin(x)} & \text{if } x \in (0, \frac{\pi}{2}] \end{cases}$. 
For $x = 0$, $\lim_{n \rightarrow \infty}f_n(0) = 0$ and 
For $x \in (0,\frac{\pi}{2}]$, $\lim_{n \rightarrow \infty}f_n(x) = \lim_{n\rightarrow \infty}\frac{x}{\sin(x) + \frac{1}{n}} = \frac{x}{\sin(x)}$ $\square$