How to find the pointwise limit of $(f_n)$ For $x \in [0, \pi/2]$, if
$$f_n(x) = \frac {nx} {1+n\sin(x)}$$
how do you find the pointwise limit of $(f_n)$ ?

Observe that the sine function obeys the bound $0\leq\sin(x)\leq 1$ for $x\in [0,\frac{\pi}{2}]$. Taking the limit of $f_n$ as $n\to\infty$ gives us
\begin{equation*}
f(x):=\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}\frac{nx}{1+n\sin(x)}=\left\{\begin{matrix}
0, & \text{for}~x=0, \\ 
\frac{x}{\sin(x)}, & \text{for}~0<x<\frac{\pi}{2}. 
\end{matrix}\right.
\end{equation*}
To get the limit for $0<x<\frac{\pi}{2}$, use either L'Hopital's rule or divide the numerator and denominator of $f_n(x)$ by $n$. 
