TITLE: If $e=\lim\limits_{x\to \infty} (1+x^{-1})^x$, why doesn't $e=1$?
QUESTION [7 upvotes]: I'm sure that this is a very basic question, but it has been bothering me for a while:
If $e=\lim\limits_{x\to \infty} (1+x^{-1})^x$, shouldn't $e=1$?
If $x$ is tending towards infinity, why wouldn't $x^{-1}$ tend to $0$ and $e=1^\infty=1$?
Thanks

REPLY [2 votes]: Intuitively, you are looking at
$$ (\text{something slightly bigger than } 1)^{\text{something going to } \infty} $$
Now, the base number is $1 + 1/n$ and it becomes closer and closer to $1$ as $n\to\infty$, with speed $n$. At the same time, the exponent tends to $\infty$ with speed $n$.
You know that for any (finite and positive) number $x$ you have $x^\infty = \infty$ and $1^x = 1$, but our situation is subtle, since the base is just tending to $1$ and the exponent is just tending to infinity. 
You kind of have two competing forces pulling a rope and the question is: who is going to win?
Surprisingly there's no winner, but an equilibrium value: the famous $e$. For this reason, I like to think that $e$ stands for equilibrium.