# Math Help - L'Hopital lim (x->0) (a^x-1)/x

1. ## L'Hopital lim (x->0) (a^x-1)/x

Good morning,

I am doing a quick review and going over some of my completed problems here. I have the following:

$
\displaystyle{\frac{\lim}{x\rightarrow0}\frac{a^x-1}{x}}
$

Applying $L'H$ I got

$
\displaystyle{\frac{\lim}{x\rightarrow0}\frac{\ln( a)a^x}{1}}
$

Here's the problem, I don't remember why I went from $a^x$to $ln(a)a^x$.

It's something simple, right?

2. $a^x = e^{\ln(a^x)}=e^{\ln(a)x}$

3. you differentiated $a^x$ which is $a^x\ln a$

just like ${d\over dx}e^x=e^x\ln e=e^x$