Limits and exponents and e exponent form So I know that $\underset{n\rightarrow \infty}{\text{lim}}\left(1+\frac {1}{n}\right)^n=e$ 
and that we're not allowed to see it as $1^\infty$ because that'd be incorrect. 
Why is then that we can do the same thing with (for example): $$\lim_{n\rightarrow \infty} \left(1+\sin\left(\frac {1}{n}\right)\right)^{n\cos\left(\frac {1}{n}\right)}= \lim_{n\rightarrow \infty} \left(\left(1+\sin\left(\frac {1}{n}\right)\right)^\frac {1}{\sin\left(\frac {1}{n}\right)} \right)^{n\cdot\cos\left(\frac {1}{n}\right)\sin\left(\frac{1}{n}\right)}$$ 
What I mean by that is that in the example of $e$ we can't say it'  $\lim \left(1+\frac {1}{n}\right)^{\lim  (n)}=1^\infty$ 
While in the second example that's exactly what we do (we say that the limit of the base is $e$ while the limit of the exponent is 1 which is why the entire expression is equal to $e$.
Can anyone explain to me the difference between the two?

In fact there's no difference between the two examples, indeed if you have a function $h$ such that $h(n)\to\infty$ then
$$\lim_{n\to\infty}\left(1+\frac{1}{h(n)}\right)^{h(n)}=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e$$
and ofcourse if we have another function $f$ such that $f(n)\to a\in\mathbb R$ then
$$\lim_{n\to\infty}\left(\left(1+\frac{1}{h(n)}\right)^{h(n)}\right)^{f(n)}=e^a.$$
