Why is $e = \lim\limits_{x\to 0} (1+x)^\frac{1}{x}$? The definition of the number $e$ that's used in my textbook is  $e = \lim\limits_{x\to \infty} (1+\frac {1}{x})^x$ which relates to compound interest. 
But when trying to calculate the derivative of $e^x$, I encountered another definition of $e$: $e = \lim\limits_{x\to 0} (1+x)^\frac{1}{x}$
Now, if it were $x$ approached $0$ from the positive direction, it would've made sense to me since $\frac {1}{x}$ wold approach infinity and $1+x$ would approached $1$ with the same "speed"(sorry I don't know the term here) as the first definition. But it's approaching $0$ from both sides.
My question is, why are these two equal?
EDIT
The reason my question isn't a duplicate of the suggested one is that mine isn't really about the limit $\lim\limits_{x\to \infty} (1+\frac {1}{x})^x$. It's about its alternative form. I totally understand how the former isn't equal to $1$.
Thank you so much in advance!
P.S: I graphed the function $(1+x)^\frac {1}{x}$ and can "see" the limit exists as $x$ approaches $0$ but I would like a non-visual proof.
 A: 
The following general result holds for any real number $y$.
  $$\lim_{n \to \infty} \left(1+\frac{y}{n}\right)^n = e^y$$

To prove this, note that the limit of the logarithm is
$$\lim_{n \to \infty}
\log((1+\frac{y}{n})^n)
=
\lim_{n \to \infty}
\frac{\log(1+y/n)}{1/n}
= \lim_{n \to \infty}\frac{y/(n^2(1+y/n))}{1/n^2}
= \lim_{n \to \infty} \frac{y}{1+y/n}
= y,$$
by l'Hôpital's rule,
so the original limit is $e^y$. Note that this assumes that you have defined $e$ somehow, and that you have defined the logarithm as the inverse of the function $x \mapsto e^x$.
If you have already shown that your definition of $e$ is equivalent to $\sum_{k = 0}^\infty \frac{1}{k!}$, then a similar argument using binomial expansions can work as well.

In particular, this result implies both $\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^x = e$
and $\lim_{x \to \infty}\left(1 - \frac{1}{x}\right)^x = e^{-1}$, which is what you need to conclude the two-sided limit $\lim_{x \to 0} (1+x)^{1/x} = e$.
A: You should look at the binomial expansion
$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... $
If you now substitute $n$ by $1/x$ you get
$(1 + x)^{1/x} = 1 + \frac{1}{x}x + \frac{\frac{1}{x}(\frac{1}{x}-1)}{2!}x^2 + \frac{\frac{1}{x}(\frac{1}{x}-1)(\frac{1}{x}-2)}{3!}x^3 + ... $
$= 1 + 1 + \frac{1(1-x)}{2!} + \frac{1(1-x)(1-2x)}{3!} + ...$
and hence
$\lim_{x\rightarrow 0}(1 + x)^{1/x} = \lim_{x\rightarrow 0} 1 + 1 + \frac{1(1-x)}{2!} + \frac{1(1-x)(1-2x)}{3!} + ...$
$ = \sum_{k = 0}^{\infty} \frac{1}{k!} = e$
A: You want to prove $$\lim_{x\to -\infty}(1+\frac1x)^x = e\\\lim_{x\to-\infty}(1+\frac1x)^x=\lim_{x\to\infty}(1-\frac1x)^{-x}=\lim_{x\to\infty}\frac{1}{(1-\frac1x)^x}=\frac1{\frac1e} = e$$
To get $$\lim_{x\to\infty}(1-\frac1x)^x = \frac1e$$
You can prove that $$\lim_{x\to\infty}(1+\frac1x)^x(1-\frac1x)^x = \lim_{x\to\infty}(1-\frac1{x^2})^x=1$$
The last limit follows because $$(1-\frac1x)\leq(1-\frac1{x^2})^x\leq1$$
Lower bound follows from Bernouli
