A Basic Limit From Exponentials Reading the proof of exponential derivatives I understand this:
To show that $(2^x)'=\ln 2 \cdot 2^x$ in the proof is used the limit:
$$\lim_{x \to 0} \frac{2^x-1}{x}$$
My question is: ¿How do I prove that this limit exist?
I don't care about its value. If I were going to prove that this limit is equal to
$\ln 2$, I would need the number $e$, and again, this number is defined as the number $a$ such that:
$$\lim_{x \to 0} \frac{a^x-1}{x} = 1$$
In another words if I know that for some constant value $a$ the limit:
$$\lim_{x \to 0} \frac{a^x-1}{x} $$
exists, let's say it's a number $L$ then I could find all the limits like these in function of $L$,
But what makes obvious that this limit exist?
 A: Let $a>0$ and $a\ne1$. We want to prove the existence of  $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$.
Assume that $r>1$ and let $f(x)=x^r-rx+r-1$ for $x>0$. Then
$$f'(x)=r(x^{r-1}-1)\begin{cases}<0 &\text{if }0<x<1\\
=0 &\text{if }x=1\\
>0 &\text{if }x>1 \end{cases}$$
Therefore, $f$ attains its absolute minimum at $x=1$. So  for all $x>0$, we have
$$f(x)\ge f(1)=0$$
$$x^r\ge rx+1-r$$
So when $r>1$ and $h>0$, $\displaystyle\frac{a^{rh}-1}{rh}\ge\frac{ra^h+1-r-1}{rh}$ and hence
\begin{align}
\frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\ge0
\end{align}
When $r>1$ and $h<0$, $\displaystyle\frac{a^{rh}-1}{rh}\le\frac{ra^h+1-r-1}{rh}$ and hence
\begin{align}
\frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\le0
\end{align}
Therefore, $\displaystyle \frac{a^h-1}{h}$ is an increasing function in $h$. As it is bounded below by $0$ on $(0,\infty)$, $\displaystyle \lim_{h\to0^+}\frac{a^h-1}{h}$ exists.
When $h<0$,
$$\frac{a^h-1}{h}=a^h\left(\frac{a^{-h}-1}{-h}\right)$$
As $\displaystyle \lim_{h\to0^-}a^h$ exists and equals $1$, $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}$ exists and $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}= \lim_{h\to0^+}\frac{a^h-1}{h}$.
Therefore, $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$ exists.
Source: How does one prove that $e$ exists?
A: $${{2^x-1}\over x}={{e^{xln(2)}-1\over x}}$$
$$\ln(2){{e^{x\ln(2)}-1\over {\ln(2)x}}}$$
Use the variable change $u=\ln(2)x$.
