# Prove the statement $\lim\limits_{h\to0}\frac{b^h-1}{h}=1 \iff b=e$.

The derivative $$\dfrac{d}{dx}\left(e^x\right)=e^x$$ can be proven a number of ways. I typically like to teach my students to prove the derivative using logarithmic differentiation: \begin{align} y&=e^x\\ \log y&=x \\ \frac{d}{dx}\left(\log y\right) &= \frac{d}{dx} \left(x\right) \\ \frac{1}{y}\cdot y'&=1 \\ y'&=y\\ y'&=e^x \end{align}

Recently I have been thinking about how to prove the derivative using the first principles definition of the derivative: $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$$ and in doing so, we need to be able to prove $$\lim_{h\to0}\frac{e^h-1}h=1.$$

I understand that one definition of $$e$$ is that it is the only number that satisfies the above limit of it is used as the base of the exponential function, and in my years of study I have chosen to simply accept it as a definition. However, I would like to be able to prove it upon request, using basic properties of limits and another definition of $$e$$: $$e=\lim_{n\to0}\left(1+n\right)^{\frac1n}.$$ Is my proof below acceptable? Is there anything about it I could improve?

Statement: $$\lim\limits_{h\to0}\frac{b^h-1}{h}=1 \iff b=e$$.

Proof. Let $$u=b^h-1$$. Then $$h=\log_b(u+1)$$, and $$h\to0\implies u\to0$$. We then have \begin{align} \lim_{h\to0}\frac{b^h-1}h&=\lim_{u\to0}\frac{u}{\log_b(u+1)}\\ &=\lim_{u\to0}\frac1{\frac1u\log_b(u+1)}\\ &=\lim_{u\to0}\frac1{\log_b(u+1)^{\frac1u}}\\ &=\frac1{\log_b\left(\lim\limits_{u\to0}(1+u)^{\frac1u}\right)}\\ \lim_{h\to0}\frac{b^h-1}h&=\frac1{\log_be} \end{align} Then, solving the equation, we have $$\frac1{\log_be}=1\iff b=e.$$

• Related article on proofwiki: proofwiki.org/wiki/Derivative_of_Exponential_at_Zero – Maximilian Janisch Feb 12 '20 at 15:39
• In $h\to0\implies u\to0$ you are hiding that $e^x$ is continuous at $x=0$ and that $e^0=1$ and a similar property for $\log_b$ when the limit switches with it. – user748968 Feb 12 '20 at 15:54
• It is quite easy to prove that $$\lim_{x\to0}\dfrac{b^x-1}{x}=\lim_{x\to0}\dfrac{b^x\log b}{1}=\log b\space \text {where the logarithm is neperian }$$ and $$\log b=1\iff b=e^1$$ – Piquito Feb 12 '20 at 15:59
• You need to justify the exchange of the limit and the logarithm. – Yves Daoust Feb 12 '20 at 16:31

Your first approach is simpler and the preferred one. It is based on the following assumption:

There exists a function $$\log:(0, \infty) \to\mathbb{R}$$ such that $$\log 1=0$$ and $$\dfrac{d} {dx} \log x=\dfrac{1}{x},\,\forall x \in(0,\infty)$$ Further the symbol $$e^x$$ is defined by $$y=e^ x\iff x=\log y$$

The above assumption is easily justified by using the definition $$\log x=\int_{1}^{x}\frac{dt}{t}$$

The second approach you have chosen is difficult. It involves defining the symbol $$a^b, a>0,b\in\mathbb {R}$$ without the use of logarithm. And then one analyzes the limit $$f(a) =\lim_{h\to 0}\frac{a^h-1}{h}$$ and shows that it exists for every $$a>0$$ and hence defines a function $$f:(0, \infty) \to\mathbb {R}$$.

Further one establishes that $$f$$ defined above is strictly increasing, continuous and the range of $$f$$ is $$\mathbb {R}$$ and $$f(1)=0,f(xy) =f(x) +f(y), f'(x) =1/x$$ Hence there is a unique number $$e>1$$ such that $$f(e) =1$$.

Once you have reached this point, it is easy to show that $$e=\lim_{n\to \infty} \left(1+\frac{1}{n}\right) ^n$$ We have $$f((1+(1/n))^n)=nf(1+(1/n))=\dfrac{f(1+(1/n))-f(1)}{1/n}\to f'(1)=1$$ as $$n\to\infty$$. Let $$g$$ be the inverse of $$f$$ so that $$g$$ is also continuous and $$g(1)=e$$. Clearly we have $$g(f((1+(1/n))^n))\to g(1)=e$$ or $$(1+(1/n))^n\to e$$ and we are done.

Hint: The derivative of $$b^x$$ is $$\ln b\cdot b^x$$.

Further hint: $$(b^x)'(0)=\lim_{h\to0}\dfrac{b^h-1}h$$.