$ \lim \frac{a^x-b^x}{x}$ as $x \to 0$ where $a>b>0$

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

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.

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 $$ \frac{a^x-b^x}{x}= \frac{e^{x\ln a}-1}{x}- \frac{e^{x\ln b}-1}{x}. $$