# Finding $\lim\limits_{x\to0+} (\frac{e^x-1}{x})^{\frac1x}$ without Taylor's series.

The solution on the textbook is $\sqrt e$. I assume I have to use the notable limit $$\lim_{x\to0+} (1+x)^{1/x}=e$$ after transforming the function into $((1+\frac x2)^{\frac 2x})^{\frac12}$ but I am stumped on the intermediate passages.

• Just use taylor expansion of $e^x$ at $x=0$ Jul 8, 2018 at 5:43
• Sorry, I should've specified in the title: no De L'Hopital and no Taylor either.
– Moss
Jul 8, 2018 at 5:44
• This is difficult without both Jul 8, 2018 at 5:50
• It is so simple with Taylor series. Jul 8, 2018 at 5:55
• No Taylor series?! What are you allowed to use? What is your definition of $e^x$? Jul 8, 2018 at 5:57

Given that $$\lim_{x\to0}\left(\frac{e^x-1}{x}\right)^{1/x}=\lim_{x\to0}\exp\left[{\frac{1}{x}\;\log\left(\frac{e^x-1}{x}\right)}\right]$$ and that $$\lim_{x\to0}\frac{1}{x}\;\log\left[1+\left(\frac{e^x-1}{x}-1\right)\right] =\lim_{x\to0}\frac{1}{x}\left(\frac{e^x-1}{x}-1\right)=\lim_{x\to0}\left(\frac{e^x-1-x}{x^2}\right)$$ this other question could be relevant:
How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion?
If you take $\log$, and use the inequality $$\frac{x-1}{x}\leq \log(x)\leq x-1$$ which is quite ideal around the value $x=1$ (which happens to be $\exp'(0)$), you can get the inequalities below: $$\frac{e^x-x-1}{x^2}\cdot \frac{x}{e^x-1}=\frac{\frac{e^x-1}{x}-1}{\frac{e^x-1}{x}}\cdot\frac{1}{x}\leq\frac{\ln\left(\frac{e^x-1}{x}\right)}{x}\leq \frac{\frac{e^x-1}{x}-1}{x}=\frac{e^x-x-1}{x^2}$$
Both functions on the left and right tend to $1/2$, if you can show those instead.