limit of ${\displaystyle \frac{a^x-1}{x}}$ as $x$ approaches 0

Corollary.

For $a\mathrm{>}\mathrm{0}$, we have

$$\underset{x\to 0}{lim}\frac{a^x-1}{x}=\ln a.$$

Proof.

Recall that $a^x=e^{x\ln a}$. Thus,

$\underset{x\to 0}{lim}{\displaystyle \frac{a^x-1}{x}}$$=\underset{x\to 0}{lim}{\displaystyle \frac{e^{x\ln a}-1}{x}}$

$=\underset{x\to 0}{lim}{\displaystyle \frac{(e^{x\ln a}-1)\ln a}{x\ln a}}$

$=(\ln a)\underset{x\to 0}{lim}{\displaystyle \frac{e^{x\ln a}-1}{x\ln a}}.$

Let $t=x\ln a$. Then $t\to 0$ as $x\to 0$. Therefore,

$\underset{x\to 0}{lim}{\displaystyle \frac{a^x-1}{x}}$$=(\ln a)\underset{t\to 0}{lim}{\displaystyle \frac{e^t-1}{t}}$

$=(\ln a)1$

$=\ln a.\mathit{∎}$

The formula from the corollary is useful for proving that ${\displaystyle \frac{d}{dx}}{a}^x=a^x\ln a$. On the other hand, once this fact is known, the corollary is easily proven via l’Hôpital’s rule (http://planetmath.org/LHpitalsRule):

$\underset{x\to 0}{lim}{\displaystyle \frac{a^x-1}{x}}$$=\underset{x\to 0}{lim}{\displaystyle \frac{a^x\ln a}{1}}$

$=a^0\ln a$

$=\ln a.$