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?

This might give you an idea about how you can generalize the limit
$$
\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = e.
$$
Suppose $a_n \to 0$ and $b_n \to \infty$.  Then
$$
\lim_{n\to\infty} (1+a_n)^{b_n} = \lim_{n\to\infty} \exp\Bigl[b_n \log(1+a_n)\Bigr] = C
$$
if and only if
$$
\lim_{n\to\infty} b_n \log(1+a_n) = \log C.
$$
Now
$$
b_n \log(1+a_n) = a_n b_n \frac{\log(1+a_n)}{a_n},
$$
and since
$$
\lim_{x \to 0} \frac{\log(1+x)}{x} = 1
$$
We may conclude that
$$
\lim_{n \to \infty} (1+a_n)^{b_n} = C
$$
if and only if
$$
\lim_{n \to \infty} a_n b_n = \log C.
$$
