$$ \lim_{x\to \:0}\left(\frac{a^x-1}{x}\right) $$
I know the answer to this is $$\ln \left(a\right)$$ but I don't know how to reach that answer without L'Hopital.
I just know the first step (given to me like a hint): use change of variable $$ a^x-1\:=\:u $$
But how when I use this I reach the non conclusive $$\lim _{u\to 0}\left(\frac{u}{\log _a\left(u+1\right)}\right)$$
or the $$\lim \:_{u\to \:0}\left(\frac{u\:\ln \left(a\right)}{\ln \left(u+1\right)}\right)$$
I know I'm doing this wrong so I would like some orientation on how to tackle problems like these.
Is there any relation to the ones solved with the definition of Euler? Those I know how to manipulate the expressions to have them look like it.
Beforehand thank you very much.
Best Answer
You've reached a perfectly conclusive point since $$\lim_{u\to 0}\frac{\ln(1+u)}u=1$$ is a high-school standard limit.
It is actually the limit of the rate of variation of the function $\ln x$, starting from $x=1$, so that its limit is just the derivative of the log at $x=1$.