Evaluate $\lim_{x\to 0} \frac{a^x -1}{x}$ without applying L'Hopital's Rule.

Hint: if you write $f(x) = a^x$, then the given limit has the form $$\lim_{x \to 0} \frac{f(x) - f(0)}{x-0}.$$ Does that look familiar?

we have

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

$$\lim_{t\to 0}\frac {e^t-1}{ t} =1$$

the limit is $$\ln(a) $$

set $$t=a^x-1$$ then we have $$x=\frac{1}{\ln(a)}\ln(t+1)$$ and you will get $$\frac{t}{\frac{1}{\ln(a)}\ln(t+1)}$$ can you finish?