TITLE: $ \lim \frac{a^x-b^x}{x}$ as $x \to 0$ where $a>b>0$
QUESTION [6 upvotes]: Calculate $ \displaystyle \lim _{x \to 0} \frac{a^x-b^x}{x}$
  where $a>b>0$

My thoughts: I think L'hopital's rule would apply. But differentiating gives me a way more complicated limit. I tried to see if it's the derivative of some function evaluated at a point, but I can't find such function.

REPLY [3 votes]: Application of L'Hospital's rule gives you:
$$  \lim _{x \to 0} \frac{a^x-b^x}{x} =  \lim _{x \to 0} a^x \cdot \log a-b^x \cdot \log b = \log a - \log b = \log {a \over b}$$

REPLY [3 votes]: Your other suggestion also works. Note that our function is equal to
$$b^x \frac{(a/b)^x-1}{x}.$$
One can recognize
$$\lim_{x\to 0}\frac{(a/b)^x-1}{x}$$
as a derivative.
Even more simply, we recognize
$$\lim_{x\to 0} \frac{a^x-b^x-0}{x}$$
as a derivarive.

REPLY [2 votes]: Hint. Let $\alpha$ be any real number. One may recall that, as $u \to 0$, one has
$$
\lim _{u \to 0} \frac{e^{\alpha u}-1}{u}=\alpha
$$ then write
$$
 \frac{a^x-b^x}{x}= \frac{e^{x\ln a}-1}{x}- \frac{e^{x\ln b}-1}{x}.
$$