Derivation for the derivative of $a^{t}$ from The Equation In Calculus, the Equation is known as:
$$f'(x)=\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
This equation allow us to find the derivatives of functions. Let's try this with the exponential function:$f(x)=a^x$ where $a \gt 1$.
$$f'(x)=\lim\limits_{h \to 0} \frac{a^{x+h}-a^x}{h}$$
$$f'(x)=\lim\limits_{h \to 0} \frac{a^x a^h-a^x}{h}$$
$$f'(x)= a^x \lim\limits_{h \to 0} \frac{a^h-1}{h}$$
As you can see, we need to determine $\lim\limits_{h \to 0} (\frac{a^h-1}{h})$ to find the derivative. At this point, we ask the question: what would $a$ be such that the limit would be 1? The answer is $e$ and that's the end of everything. 
What I want to know is can we actually evaluate that limit? How can we actually find $e$ other than guessing with trial and error?
 A: You got the expression 
$$a^x \lim\limits_{h \to 0} \frac{a^h-1}{h}.$$
We make the change of variables $h=\log_a x$ (see that $a>1$). Then
$$
\lim\limits_{h \to 0} \frac{a^h-1}{h}=\log a\lim_{x\to 1}\frac{x-1}{\log x}=\log a\frac{1}{\lim_{x\to 1}\frac{\log x}{x-1}}.
$$
Now, this limit is equal to
$$
\lim_{x\to 1}\frac{\log x}{x-1}=\lim_{x\to 1}\log(x^{1/(x-1)}).
$$
Doing the change $x=t+1$ we arrive to
$$
\lim_{x\to 1}\frac{\log x}{x-1}=\lim_{t\to 0}\log(t+1)^{1/t}=1.
$$
A: One of the several definitions of $e$ is:
$$e = \lim_{x \to 0} (1 + x)^{1/x}.$$
Taking the natural log of both sides and interchanging the log and the limit (which is allowed by definition of continuity),
$$1 = \ln\left(\lim_{x \to 0} (1 + x)^{1/x}\right) = \lim_{x \to 0}\left( \ln (1 + x)^{1/x}\right) = \lim_{x \to 0} \frac{\ln(1 + x)}{x} = \lim_{x \to 0} \frac{x}{\ln(1 + x)}$$
Let $y = a^h - 1$. Then $h = \frac{\ln(y+1)}{\ln a}$, so
$$\lim_{h \to 0} \frac{a^h - 1}{h} = \lim_{y \to 0} \frac{y \ln a}{\ln(y+1)} = \ln a \cdot \lim_{y \to 0} \frac{y}{\ln(y+1)} = \ln a.$$
A: You can use binom theory to find taylor expension result of $$ \lim\limits_{h \to 0} \frac{a^h-1}{h}$$
if $a=k+1$
$(1+k)^h=1+C(h,1)k+C(h,2)k^2+C(h,3)k^3+.....$
$(1+k)^h=1+hk+\frac{h(h-1)}{2!}k^2+\frac{h(h-1)(h-2)}{3!}k^3+.....$
$$g(k)= \lim\limits_{h \to 0} \frac{(1+k)^{h}-1}{h}=\frac{hk+\frac{h(h-1)}{2!}k^2+\frac{h(h-1)(h-2)}{3!}k^3+.....}{h}=\lim\limits_{h \to 0} (k-\frac{k^2}{2}+\frac{k^3}{3}-\frac{k^4}{4}+....)+hU_1(k)+h^2U_2(k)+....$$
$$ \lim\limits_{h \to 0} \frac{(1+k)^{h}-1}{h}=k-\frac{k^2}{2}+\frac{k^3}{3}-\frac{k^4}{4}+....$$
$$ g(k)=k-\frac{k^2}{2}+\frac{k^3}{3}-\frac{k^4}{4}+....$$
$$ g(a-1)=\lim\limits_{h \to 0} \frac{a^h-1}{h}$$
$$ g(b-1)=\lim\limits_{h \to 0} \frac{b^h-1}{h}$$
$$ g(ab-1)=\lim\limits_{h \to 0} \frac{(ab)^h-1}{h}=\lim\limits_{h \to 0} \frac{(ab)^h-a^h+a^h-1}{h}=\lim\limits_{h \to 0} \frac{a^h(b^h-1)}{h}+\lim\limits_{h \to 0} \frac{a^h-1}{h}=\lim\limits_{h \to 0} \frac{(b^h-1)}{h}+\lim\limits_{h \to 0} \frac{a^h-1}{h}=g(b-1)+g(a-1)$$
We have a relation for $g(x)$ function.
$$ g(ab-1)=g(b-1)+g(a-1)$$
and if you put $b=1/a$
$$ g(a^{-1}-1)=-g(a-1)$$
It is easy to show $g(0)=0$   
you want to find $g(e)=1$  thus
First try to find $g(1)$  ($ g(1)= \lim\limits_{h \to 0} \frac{2^h-1}{h}$
$$ g(1)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....$$
even you calculate first 4 terms it will be near to $0.58$ (it is about 0.69)
$g(1)\approx 0.69$
$$ g(2.2-1)=g(1)+g(1)$$
$$ g(3)=g(1)+g(1)\approx 1.38$$
thus we can see that $2<e<4$
$$g(1/2)=\frac{1}{2}-\frac{1}{4.2}+\frac{1}{8.3}-\frac{1}{16.4}+....$$
$$g(1/2)\approx  0.4$$
To use $$ g(ab-1)=g(b-1)+g(a-1)$$  again
$$ g(3.3/2.2 -1)=g(3/2-1)+g(3/2-1)$$
$$ g(9/4 -1)=g(3/2-1)+g(3/2-1)$$
$$ g(9/4 -1)=2g(1/2)$$
$$ g(5/4 )=2g(1/2) \approx 0.8$$
It means $e>5/4+1=2.25$
You can try more values to get near to e value.
You can use with such Technics to find e value range
A: Notice, $$a^h=1+\frac{h}{1!}(\log a)+\frac{h^2}{2!}(\log a)^2+\frac{h^3}{3!}(\log a)^3+\ldots$$ Now, we have $$\lim_{h\to 0}\frac{a^h-1}{h}$$
$$=\lim_{h\to 0}\frac{\left(1+\frac{h}{1!}(\log a)+\frac{h^2}{2!}(\log a)^2+\frac{h^3}{3!}(\log a)^3+\ldots\right)-1}{h}$$
$$=\lim_{h\to 0}\frac{\left(\frac{h}{1!}(\log a)+\frac{h^2}{2!}(\log a)^2+\frac{h^3}{3!}(\log a)^3+\ldots\right)}{h}$$
$$=\lim_{h\to 0}\left(\frac{1}{1!}(\log a)+\frac{h}{2!}(\log a)^2+\frac{h^2}{3!}(\log a)^3+\ldots\right)$$
$$=\lim_{h\to 0}\left(\frac{1}{1!}(\log a)+0\right)=\log a$$
Now, the limit will be $1$ if we have $$\log a=1$$$$\iff a=e^1=e$$
A: From
$f'(x)= a^x \lim\limits_{h \to 0} \frac{a^h-1}{h}
$,
$f'(x)
=a^x f'(0)
$,
since $f(0) = 1$,
so
$\lim\limits_{h \to 0} \frac{a^h-1}{h}
=\lim\limits_{h \to 0} \frac{f(h)-f(0)}{h}
=f'(0)
$.
The "natural" solution
is the one where
$f'(0) = 1$,
and the other answers show
a variety of ways
to conclude that
the value that does this is
$a=e
=\lim_{n \to \infty} (1+\frac1{n})^n
=\lim_{x \to 0} (1+x)^{1/x}
$.
A: Yet another way is to evaluate the limit of interest is to invoke the definition for $e^x$ expressed as 
$$e^x=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n \tag 1$$
Using $(1)$, we can write
$$\lim_{h\to 0}\frac{a^h-1}{h}=\lim_{h\to 0}\lim_{n\to \infty}\frac{\left(1+\frac{h\log a}{n}\right)^n-1}{h} \tag 2$$
Next, we recall from the Mean Value Theorem that there exists a number $\xi$, ($0<\xi<h\log a/n$ for $a>1$,  $h\log a/n<\xi<0$, for $0<a<1$),such that
$$\left(1+\frac{h\log a}{n}\right)^n=1+h\,\log a+\frac{n(n-1)}{2n^2}(1+\xi)^{n-2} (h\log a)^2 \tag 3$$
Therefore, using $(3)$ in $(2)$ reveals 
$$\begin{align}
\lim_{h\to 0}\lim_{n\to \infty}\frac{\left(1+\frac{h\log a}{n}\right)^n-1}{h}&=\log a+\lim_{h\to 0}\lim_{n\to \infty}\frac{n(n-1)}{2n^2}(1+\xi)^{n-2} h\log^2 a\\\\
&=\log a
\end{align}$$
as was to be shown!
