# How to prove that $\lim\limits_{h \to 0} \frac{a^h - 1}{h} = \ln a$

In order to find the derivative of a exponential function, on its general form $a^x$ by the definition, I used limits.

\begin{align*} \frac{d}{dx} a^x & = \lim_{h \to 0} \left [ \frac{a^{x+h}-a^x}{h} \right ]\\ \\ & =\lim_{h \to 0} \left [ \frac{a^x \cdot a^h-a^x}{h} \right ] \\ \\ &=\lim_{h \to 0} \left [ \frac{a^x \cdot (a^h-1)}{h} \right ] \\ \\ &=a^x \cdot \lim_{h \to 0} \left [\frac {a^h-1}{h} \right ] \end{align*}

I know that this last limit is equal to $\ln(a)$ but how can I prove it by using basic Algebra and Exponential and Logarithms properties? Thanks

• $a^h=\exp(\ln a \cdot h)$. Are you able to Taylor expand the exponential function? – anon Oct 30 '11 at 23:33
• Well, can we use the definition but be a bit witty? Perhaps we could use the definition (along with the chain rule) on $\log y = \log (a^x)$? – davidlowryduda Oct 30 '11 at 23:43
• I think it is worth mentioning that some authors (such as E. Landau) define the logarithm that way. – Pedro Tamaroff May 2 '12 at 21:23
• I consider this to be a rather tricky issue. Let me explain. Let $a$ be positive real number. Then to \textit{define} $a^x$ algebraically is easy on the rationals, just ask that the standard exponent laws hold. But if you want to define $a^x$ for irrational values, you need to do some analysis. In particular you need to prove that $a^x$ is continuous on the rationals. – Baby Dragon Jun 3 '12 at 8:04
• math.stackexchange.com/questions/1828962/… – user301988 Jul 4 '16 at 0:25

First, we prove it for $a=e$.
Let $\frac{1}{t} = e^h - 1$. Then $e^h = 1+\frac{1}{t}$, so $h = \ln(1 + \frac{1}{t})$. As $h\to 0$, we have $\frac{1}{t}\to 0$, so $t\to \infty$ (if $h\to 0^+$) or $t\to-\infty$ (if $h\to 0^-$). We have: \begin{align*} \lim_{h\to 0^+}\frac{e^h-1}{h} &= \lim_{t\to\infty}\left(\frac{1/t}{\ln(1+\frac{1}{t})}\right)\\ &= \lim_{t\to\infty}\frac{1}{t\ln(1+\frac{1}{t})}\\ &= \lim_{t\to\infty}\frac{1}{\ln\left( (1+\frac{1}{t})^t\right)}\\ &= \frac{1}{\ln\left(\lim_{t\to\infty}(1 + \frac{1}{t})^t\right)}\\ &= \frac{1}{\ln(e)}\\ &= 1. \end{align*} Since we also have \begin{align*} \lim_{t\to-\infty}\left(1 + \frac{1}{t}\right)^t &= \lim_{n\to\infty}\left(1 - \frac{1}{n}\right)^{-n}\\ &= \lim_{n\to\infty}\frac{1}{(1 + \frac{-1}{n})^n}\\ &= \frac{1}{e^{-1}}\\ &= e,\end{align*} a similar calculation shows that $$\lim_{h\to 0^-}\frac{e^h-1}{h} = 1.$$ Hence, $$\lim_{h\to 0}\frac{e^h-1}{h} = 1.$$
Now for arbitrary $a\gt 0$, $a\neq 1$, we rewrite $a^h = e^{h\ln(a)}$, and make the substitution $t = h\ln(a)$. Then the result follows from the case $a=h$ as Gerry Myerson shows.
It depends a bit on what you're prepared to accept as "basic algebra and exponential and logarithms properties". Look first at the case where $a$ is $e$. You need to know that $\lim_{h\to0}(e^h-1)/h)=1$. Are you willing to accept that as a "basic property"? If so, then $a^h=e^{h\log a}$ so $$(a^h-1)/h=(e^{h\log a}-1)/h={e^{h\log a}-1\over h\log a}\log a$$ so $$\lim_{h\to0}(a^h-1)/h=(\log a)\lim_{h\to0}{e^{h\log a}-1\over h\log a}=\log a$$