# 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?

• I'd argue that in some sense, it isn't obvious that the limit exists. It requires proof. If you have L'Hopital's rule and basic properties of the exponential function that is one way to do it. If you have a power series definition of $e^x$ and facts about power series that is another way. If you don't have these things (and just have the definition of real exponentiation $a^b$ in terms of suprema of some set, or something similar) you may have to do some fiddling with inequalities (which may be equivalent to establishing properties of the exponential function $e^x$ or the number $e$). Feb 21, 2021 at 16:08
• Before proving the limit, you need to know what definition of $2^x$ you use.// What book are you reading?
– user9464
Feb 21, 2021 at 16:09
• Thanks for the ideas. I'm learning by 3Blue1Brown Videos (very nice) and a calculus book called Stewart. ¿Could you suggest me a book? Feb 21, 2021 at 16:12
• @user232560: the way Stewart proves the derivative of $b^x$ is via the chain rule and several other statements. Where are you exactly "Reading the proof of exponential derivatives"? In the book or somewhere else?
– user9464
Feb 21, 2021 at 16:23
• I disagree with the closing of this question. The question asked it not even similar to the one in the suggested duplicate. Feb 21, 2021 at 16:33

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

$$f'(x)=r(x^{r-1}-1)\begin{cases}<0 &\text{if }00 &\text{if }x>1 \end{cases}$$

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.

• vitamin d, That's the proof i was looking for. Feb 21, 2021 at 16:17

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

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

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

• Yes, that's the idea but that metod uses that: Feb 21, 2021 at 16:15