Intuitively, why are the two limit definitions of $e^x$ equivalent? Thanks for reading!
Intuitively, why does...
$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{xn}=\lim_{n \rightarrow \infty} \left(1+\frac{x}{n}\right)^{n}=e^x$$
Note, I'm not asking why $e^x$ is one of the two limits. I understand the first limit, or at least I think I do.
In terms of continuous growth however (I don't just want a mathematical proof - the more intuitive the answer is the better), I'd like to understand why the two limits are equivalent!
Why is letting some principal amount grow by $\frac{1}{n}$ times its current value $xn$ times equal to letting that principal grow by $\frac{x}{n}$ times its current value $n$ times if we allow $n \rightarrow \infty$?
Thanks!
 A: In general
$$\lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx}=e^{ab}$$
and they are only based on the definition of the number $e$
$$e:=\lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^n $$
Here is a simple proof: 
$$\begin{align} 
\lim_{x\to \infty} \left( 1+\frac{a}{x} \right) ^{bx} 
&= \lim_{x\to \infty} \left[ \left( 1+\frac{a}{x} \right)^{x/a} \right]^{ab} \\
&= \lim_{n\to \infty} \left[ \left( 1+\frac{1}{n} \right)^n \right]^{ab} \quad (\textrm{take } n=x/a) \\
&=  \left[ \lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^n \right]^{ab} \\
&= e^{ab}.
\end{align}$$
A: Apply binomial theorem resp. the binomial series, then
$$
\left(1+\frac1n\right)^{nx}=1+x+\sum_{k=2}^{\infty}\frac{x(x-\frac1n)...(x-\frac{k-1}n)}{k!}
$$
and 
$$
\left(1+\frac xn\right)^{n}=1+x+\sum_{k=2}^{\infty}\frac{(1-\frac1n)...(1-\frac{k-1}n)}{k!}x^k
$$
Both expansions converge in their coefficients to the same limit $\frac{x^k}{k!}$, the complicated part is to show that exchanging the limit in the series and in the coefficients is permissible.
A: Consider that for $x>0$, $n$ and $\frac nx$ both tend to infinity so that
$$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=\lim_{n\to\infty}\left(1+\dfrac xn\right)^{n/x}=e.$$
Then by continuity, you can raise to the $x^{th}$ power inside the limit,
$$\lim_{n\to\infty}\left(1+\dfrac1n\right)^{nx}=\lim_{n\to\infty}\left(1+\dfrac xn\right)^n=e^x.$$

For negative $x$, you must modify the reasoning as the argument goes to minus infinity.
$$\lim_{n\to\infty}\left(1+\dfrac xn\right)^{n/x}=\lim_{n\to\infty}\left(1-\dfrac1n\right)^{-n}=\lim_{n\to\infty}\left(1-\dfrac1{n+1}\right)^{-1-n}=\lim_{n\to\infty}\left(1+\dfrac1n\right)^{n+1}\\=\lim_{n\to\infty}\left(1+\dfrac1n\right)^n\left(1+\dfrac1n\right)=\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e.$$

For a more intuitive approach, notice that for small $\epsilon$ you can linearize
$$(1+\epsilon)^x\approx 1+\epsilon x$$ so that for large $n$,
$$\left(1+\frac1n\right)^x\approx1+\frac xn$$ and
$$\left(1+\frac1n\right)^{nx}\approx\left(1+\frac xn\right)^n.$$
For larger and larger $n$, the approximation improves.
E.g.
$$1.001^3=1.003003001\approx 1.003$$
and
$$1.001^{3000}=20.05545\approx 1.003^{1000}=19.99553\approx e^3=20.08553$$
Next,
$$1.0001^3=1.000300030001\approx 1.0003$$
and
$$1.0001^{30000}=20.08252\approx 1.0003^{10000}=20.07650\approx e^3=20.08553$$
$$\cdots$$
A: Forget that $n$ is an integer. For $x > 0$ we have
$$\lim_{y\to\infty} \left(1+\frac{x}{y}\right)^y = \lim_{y\to\infty} \left(1+\frac{1}{\frac{y}{x}}\right)^{x\cdot \frac{y}{x}}$$
If $y \to \infty$ then $z:= \frac{y}{x} \to \infty$ as well so change of variables gives that this is equal to $$\lim_{z\to\infty} \left(1+\frac1z\right)^{xz}$$
which is the desired expression.
A: I will say my first thought. 
If fix $x$,  one can take $\exp (x)$ for "constant" intuitively. And 
 it is known that for a real number $x'$ there are infinite sequences convergent to it, I mean there are many many sequences we do not construct (or notice) with the same limit $x'$, 
so let $a_n = (1+\frac{x}{n})^n$, $b_n = (1+\frac{1}{n})^{xn}$, sequences $\{ a_n \} ,  \{ b_n \}$  are just two cases( we are familiar with ) among them. 
Maybe there is nothing special comparing to the rest. 
