Limit of exponential function using natural log and change of variable without L'Hopital

Question

$$ \lim_{x\to \:0}\left(\frac{a^x-1}{x}\right) $$ 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.

Answer 1

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 $x=1$.

Answer 2

We can use here the definition of $e$.

Indeed, since $\ln$ is a continuous function we obtain: $$\lim_{u\rightarrow0}\frac{u}{\ln(1+u)}=\lim_{u\rightarrow0}\frac{1}{\ln(1+u)^{\frac{1}{u}}}=\frac{1}{\ln\lim\limits_{u\rightarrow0}(1+u)^{\frac{1}{u}}}=\frac{1}{\ln{e}}=1.$$

Answer 3

What you did is good.

I do not know if, in this context, limit-without-lhopital excludes Taylor series. If not, you could have more than the limit itself writing $$a^x=e^{x \log(a)}=1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+O\left(x^3\right)$$ which makes $$\frac{a^x-1}{x}=\log (a)+\frac{1}{2} x \log ^2(a)+O\left(x^2\right)$$ which shows the limit and also how it is approached.

Let for example $a=\pi$ and $x=\frac 1 {10}$. The exact value $10 \left(\sqrt[10]{\pi }-1\right)\approx 1.21282$ while the above truncated expression would give $\frac{1}{20} \log (\pi ) (20+\log (\pi ))\approx 1.21025$.