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 $$\frac1{\log_be}=1\iff b=e.$$

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    $\begingroup$ Related article on proofwiki: proofwiki.org/wiki/Derivative_of_Exponential_at_Zero $\endgroup$ Commented Feb 12, 2020 at 15:39
  • $\begingroup$ 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. $\endgroup$
    – user748968
    Commented Feb 12, 2020 at 15:54
  • $\begingroup$ 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 $$\log b=1\iff b=e^1$$ $\endgroup$
    – Piquito
    Commented Feb 12, 2020 at 15:59
  • $\begingroup$ You need to justify the exchange of the limit and the logarithm. $\endgroup$
    – user65203
    Commented Feb 12, 2020 at 16:31

2 Answers 2


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


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