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

###### Corollary.

For $a>0$, we have

 $\lim_{x\to 0}\frac{a^{x}-1}{x}=\ln a.$
###### Proof.

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

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

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

 $\displaystyle\lim_{x\to 0}\frac{a^{x}-1}{x}$ $\displaystyle=(\ln a)\lim_{t\to 0}\frac{e^{t}-1}{t}$ $\displaystyle=(\ln a)1$ $\displaystyle=\ln a.\qed$

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):

 $\displaystyle\lim_{x\to 0}\frac{a^{x}-1}{x}$ $\displaystyle=\lim_{x\to 0}\frac{a^{x}\ln a}{1}$ $\displaystyle=a^{0}\ln a$ $\displaystyle=\ln a.$
Title limit of $\displaystyle\frac{a^{x}-1}{x}$ as $x$ approaches 0 LimitOfdisplaystylefracax1xAsXApproaches0 2013-03-22 17:40:21 2013-03-22 17:40:21 Wkbj79 (1863) Wkbj79 (1863) 5 Wkbj79 (1863) Corollary msc 32A05