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: 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.
A: 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 $$\lim_{h\to0}(a^h-1)/h=(\log a)\lim_{h\to0}{e^{h\log a}-1\over h\log a}=\log a$$
