I was looking in into my calculus textbook and the author was proving that the derivative of $e^x$ is $e^x$ using limits and there comes a stage where the author has to prove the following limits: $$\begin{align*} \lim_{x\rightarrow 0} \frac{e^x−1}{x} = 1 \end{align*}$$
According to his technique, let $y=e^x−1$ as $y\rightarrow 0$, then $x = \ln(y + 1)$. So the limit becomes: $$\begin{align*} \lim_{y\rightarrow 0} \frac{y}{\ln{(y + 1)}} & = \frac{1}{\ln(y + 1)^\frac{1}{y}}\\ & = 1 \end{align*}$$
Why is $\frac{1}{\ln(1+y)^\frac{1}{y}}$ as $y\rightarrow 0$ equal to $1$?
Can I get a proof without L'Hôpital's rule?
It’s just a definition of $e$: $$\lim_{y\to 0} (1+y)^{1/y}= e $$ and so $$\frac{1}{\ln (1+y)^{1/y} }\to \frac{1}{\ln e} =1 $$