Showing Uniform convergence of $\frac{n x}{1 + n \sin(x)}$ I want to prove for all $a\in \left(0,\frac{\pi}{2}\right]$, $ \ f_n\to f$ uniformly on $\left[a,\frac{\pi}{2}\right]$.
Also, how is this different from $f_n \to f$ uniformly on $\left(0, \frac{\pi}{2}\right]$ ?
I made the crucial error of omitting that first line, when I asked this question before.

Let us define:
$$f_n(x) = \frac{n x}{1 + n \sin(x)}$$
Which has pointwise convergence to $f(x)$ if:
$$x = 0, \ f(x) = 0, \text{ and } x \in \left(0, \frac{\pi}{2}\,\right], \ f(x) = \frac{x}{\sin(x)}$$

Here is my attempt at the problem:
If $x \in \left(0, \frac{\pi}{2}\right]$ then 
$$\left|\, f_n(x) - f(x) \right|  = 
\left| \frac{nx}{1 + n\sin(x)} - \frac{x}{\sin(x)}\right| 
\\
= \left|\frac{nx \sin(x) - x\, \big(1 + n \sin(x)\big)}{\big(1+ n \sin(x)\big)\sin(x)} \right|
\\
 = \frac{x}{\sin(x) + n \sin^2(x)} \leq \frac{1}{n},
$$
is this line correct?
So $\forall \epsilon > 0$, we may choose $N \geq \frac{1}{\epsilon}$ such that when $n \geq N \implies \left|\,f_n(x)-f(x)\right| \leq \epsilon \quad \forall x \in \left(0, \frac{\pi}{2}\right]$

I am not satisfied or confident in my answer, may anyone else suggest improvements?

The non-uniform convergence on $(0,\pi/2]$ was addressed (partially) in another post. It follows from $\displaystyle \lim_{n \to \infty}|f_n(x) - f(x)|= 1$ for any sequence $(x_n)$ with $x_n \to 0$.
Convergence is uniform on $[a,\pi/2]$ for $0 < a  < \pi/2$.  
Note that as $n \to \infty$ we have for all $x \in [a,\pi/2]$,
$$|f_n(x) - f(x)| = \frac{x}{\sin x + n \sin^2 x} \leqslant \frac{\pi}{2n \sin^2 a}\to 0.$$
