How to visualize Euler's number? I am interested if there is geometric meaning (using graphs) of $(1 + \frac{1}{n})^n$ when $n \rightarrow \infty$. Also, is there visual explanation of why is $e^x = (1 + \frac{x}{n})^n$ when $n \rightarrow \infty$ and why is $\frac{d}{dx}e^x = e^x$?
I see that this kind of question is not posted yet.
 A: Exponentiation turns addition to product, $$a^{b+c}=a^ba^c$$ (in the naturals, this is immediate from the definition). This corresponds to a "translation" property: shifting the argument amounts to a multiplication by a constant, and conversely, multiplying by a constant preserves the shape.
By definition, the slope of a curve is the vertical increment corresponding to an horizontal increment, and by the above property, the vertical increment must be a constant times the function. Hence the derivative of an exponential is an exponential.
More specifically,
$$\lim_{h\to0}\frac{a^{x+h}-a^x}h=a^x\lim_{x\to h}\frac{a^h-1}h$$ confirms this intuition.
Now we have this "magical" number $$\lim_{x\to h}\frac{a^h-1}h,$$ which is a priori a function of $a$. When $a=1$, this is $0$; when $a=10$ (say), numerical estimates based on $h=2^{-k}$ are $9,
4.32\cdots,3.11\cdots,2.67\cdots,2.48\cdots,\cdots2.3025\cdots$. They seem to stabilize above $1$.
It is possible (I will not attempt here) to show that the limit indeed converges to a value above $1$ for $a=10$, and that it is a continuous function of $a$. Hence, by the IVT, the must exist a constant, let $e$, such that
$$\lim_{h\to0}\frac{e^h-1}h=1$$
and
$$(e^x)'=e^x.$$

The plot below illustrates the relation between an exponential and its derivative, by showing
$$3^x,3^{x+1}-3^x,3^{x+1}.$$


One can also show that
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$ is truly a power function and
$$e^x=\left(\lim_{m\to\infty}\left(1+\frac1m\right)^m\right)^x$$ (it suffices to substitute $mx$ for $n$), and the definition of the natural exponential just rests on the constant $e$.
A: I think of my favorite, and pretty geometric, proof of this limit, using the squeeze or sandwich theorem for limits.  You can do it using an upper and lower Riemann sum with one subdivision for the integral of $1/t$.
One has $L\le\int_1^{1+x/n}1/t\rm dt\le U\implies x/n(1/(1+x/n))\le\ln(1+x/n)\le x/n(1)\implies x/(n+x)\le\ln(1+x/n)\le x/n\implies e^{x/(n+x)}\le(1+x/n)\le e^{x/n}\implies e^{nx/(n+x)}\le(l+x/n)^n\le e^x$, and take limits.
