TITLE: Why isn't $\lim \limits_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}$ equal to $1$?
QUESTION [21 upvotes]: Given $\lim \limits_{x\to\infty}(1+\frac{1}{x})^{x}$, why can't you reduce it to $\lim \limits_{x\to\infty}(1+0)^{x}$, making the result "$1$"?  Obviously, it's wrong, as the true value is $e$.  Is it because the $\frac{1}{x}$ is still something even though it's really small?  Then why is $$\lim_{x\to\infty}\left(\frac{1}{x}\right) = 0\text{?}$$
What is the proper way of calculating the limit in this case?

REPLY [4 votes]: Answer to your title question. It looks like this argument has not been given yet. Except by robjohn, but as a consequence of a more complicated fact.
If this was true, then we would have in particular
$$
\lim_{n\rightarrow+\infty}\left(1+\frac{1}{n}\right)^n=1
$$
along $\mathbb{N}^*$. But for positive integers, the binomial theorem yields

$$
\left(1+\frac{1}{n}\right)^n=1+n\frac{1}{n}+\frac{n(n-1)}{2}\frac{1}{n^2}+\ldots\geq 1+1=2\qquad\forall n\geq 1.
$$