Evaluating $\lim_{n \to +\infty}\frac{\left(1 + \sin\alpha\right)^n}{\left(1+\frac{\sin\alpha}n\right)^n}$ 
Let, for $n \ge 2$,
  $$a_n = \frac{\left(1 + \sin\alpha\right)^n}{\left(1+\frac{\sin\alpha}n\right)^n}$$
  Evaluate $\lim_\limits{n \to +\infty}a_n$, as $\alpha$ varies in $[0, 2\pi)$.

I proceeded as follows.
The denominator is $\exp(\sin\alpha)$, and we can rewrite the numerator as $\exp(n\ln(1 + \sin\alpha))$. Therefore
$$a_n = \exp(n\ln(1 + \sin\alpha) - \sin\alpha)$$
So
$$\lim_{n \to +\infty} a_n = \begin{cases}0\qquad&\alpha \in \{0, \pi\}\\
+\infty\qquad&\alpha\in(0, \pi)\cup(\pi, 3\pi/2)\cup(3\pi/2,2\pi)\\
??\qquad&\alpha = 3\pi/2
\end{cases}$$
I don't know what happens when $\alpha=3\pi/2\implies\sin\alpha = -1$ because that's outside the domain of the logarithm. But $\alpha=3\pi/2$ is inside the domain of the original $a_n$ so I think I've made a mistake somewhere.

Your first step is correct, with the simpler denominator.
It was a good idea to convert to exp and log.
Also to break into quadrants with particular interest in multiples of $\pi/2$.
But then, when $x=0,\pi$, you have exp(n(0)-0) = exp(0)=1.
When $\sin x>0$, then $1+\sin x>1$, so $\ln(1+\sin x)>0$ and $n\ln(1+\sin x)\to\infty$.
Check $\sin x<0$.  For $x=3\pi/2$, go back to the original question, where it is particularly simple.  Logs won't work in this case, as you found.
