In deriving the derivative of function $f(x)=\exp(x)$, it is often pointed out that in the general case of $f_a(x)=a^x$ the following expression can be deduced from the definition of the derivative:

$$ \frac{d}{dx}a^x = a^x\cdot\left(\lim_{h\rightarrow0}\frac{a^h-1}{h}\right) $$

with $e$ defined as the real number for which the above limit term is equal to 1.

However, this seems unsatisfactory to me as the existence of a real number $a$ satisfying the equation $\lim_{h\rightarrow0}\frac{a^h-1}{h}=1$ is not established.

Can somebody please point me to a proof of the existence of such a number or provide it if possible?


  • 2
    $\begingroup$ Another problem: how do you define $a^h$ for real $a>0$ and $h$ without the exponential and the logarithm? It's feasible, for instance with Dedekind cuts, but not obvious. $\endgroup$ Feb 12 '20 at 22:31
  • $\begingroup$ Simple approach $a^h=e^{hln(a)}\approx 1+hln(a)$, so the limit $=ln(a)$. $\endgroup$ Feb 12 '20 at 22:44
  • 1
    $\begingroup$ This is already handled here : math.stackexchange.com/a/1735035/72031 $\endgroup$
    – Paramanand Singh
    Feb 13 '20 at 1:10

Assuming that basic theorems about limits and the logarithm function are available as well as $\lim_{x\to 0}\left(x+1\right)^{1/x}=e$, we can first prove that: $$\lim_{y\to 0}\frac{\ln(y+1)}{y}=\lim_{y\to 0}\ln\left(y+1\right)^{1/y}=\ln\left(\lim_{y\to 0}\left(y+1\right)^{1/y}\right)=\ln(e)=1 $$ Switching the limit and the logarithm is possible since $\ln(t)$ is continous. Then $$\lim_{y\to 0}\frac{\log_a(y+1)}{y}=\frac{1}{\ln(a)}\lim_{y\to 0}\frac{\ln(y+1)}{y}=\frac{1}{\ln(a)} $$ And $y=a^{h}-1\to 0$ as $h\to 0$. So by limit of a composition of functions, $$\lim_{h\to 0}\frac{\log_a\left(a^h-1+1\right)}{a^h-1}=\lim_{h\to 0}\frac{h}{a^h-1}=\frac{1}{\ln(a)}\\ \lim_{h\to 0}\frac{a^h-1}{h}=\ln(a)$$ Hence $\lim_{h\to 0}\dfrac{a^h-1}{h}=1$ if and only if $\ln(a)=1\Leftrightarrow a=e$.


One possible approach is to define $e$ as the limit of the sequence $$(1+\frac{1}{n})^n$$ It takes a bit of work to show that the above sequence is increasing, and not hard to show that is bounded above by $3$. With some more work one shows that $$(1+\frac{1}{n})^{n+1}$$ is decreasing. They both have the same limit, which one defines to be the number $e$.

From here one checks that $$e^{\frac{1}{n+1}}<1+\frac{1}{n}<e^{\frac{1}{n}}$$ and this implies $$\lim_{x\to 0_{+}}\frac{e^x-1}{x}=1$$ and from this $$\lim_{x\to 0}\frac{e^x-1}{x}=1$$


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