TITLE: Why $\lim \limits_ {n\to \infty}\left (\frac{n+3}{n+4}\right)^n \neq 1$?
QUESTION [19 upvotes]: Why doesn't $\lim\limits_ {n\to \infty}\ (\frac{n+3}{n+4})^n$ equal $1$? 

So this is the question.
I found it actually it equals $e^{-1}$. I could prove it, using some reordering and canceling. 
However another way I took was this:
$$\lim_ {n\to \infty}\ \left(\frac{n}{n+4}+\frac{3}{n+4}\right)^n$$
with the limit of the first term going to $1$ and the second to $0$. So $(1+0)^n=1$ not $e^{-1}$.

REPLY [2 votes]: While intuitively it seems "obvious" that $1^\infty$ has to be one, let me point to you that $ \dfrac{n+3}{n+4} <1$. Then the meaning of 
$$\left(\frac{n+3}{n+4}\right)^n $$ 
is a number strictly less than $1$ to a huge power. Now, a number between $0$ and $1$ raised to a larger power gets smaller and smaller, so is it still obvious that it approaches 1?