A Basic Limit From Exponentials

Question

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?

Answer 1

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?

Answer 2

$${{2^x-1}\over x}={{e^{xln(2)}-1\over x}}$$

$$\ln(2){{e^{x\ln(2)}-1\over {\ln(2)x}}}$$

Use the variable change $u=\ln(2)x$.