$\lim_{x \to 0}\left( \dfrac{e^x-1}x \right)^\frac1x=\lim_{x \to 0}\left( \dfrac{(1+x+\frac{x^2}2+o(x^2))-1}x \right)^\frac1x=\lim_{x \to 0}\left( {1+\frac{x}2+o(x)} \right)^\frac1x=\lim_{x \to 0}e^{\frac1x \ln\left( {1+\frac{x}2+o(x)} \right)}=\lim_{x \to 0}e^{\frac1x \left( {\frac{x}2+o(x)} \right)}=\sqrt e$
A good trick for $1^\infty$ indeterminate forms that may be worth rembering is, if $\lim f(x) = 1$ and $\lim g(x) = \infty$ then
$$\lim f(x)^{g(x)} = \exp[\lim g(x)(f(x)-1)].$$
In this case
$$\lim_{x \to 0}\left(\frac{e^x-1}x\right)^{1/x}=\exp\left(\lim_{x \to 0}\frac{e^x-1-x}{x^2}\right).$$
Using L'Hospitals rule
$$\lim_{x \to 0}\frac{e^x-1-x}{x^2}= \lim_{x \to 0}\frac{e^x-1}{2x}= \lim_{x \to 0}\frac{e^x}{2}= \frac1{2},$$
Therefore,
$$\lim_{x \to 0}\left(\frac{e^x-1}x\right)^{1/x}= e^{1/2}.$$
The proof of the trick rests on the well-known inequality $x/(1+x) \leqslant \ln(1+x) \leqslant x$ for $x > -1$.
Thus,
$$\frac{g(x)(f(x) - 1)}{f(x)} \leqslant g(x) \ln f(x) = g(x) \ln(1 + f(x) -1) \leqslant g(x)(f(x) - 1).$$
By the squeeze theorem
$$\lim g(x) \ln f(x) = \lim g(x)( f(x) -1).$$
The first thing I would do is take the logarithm of the whole thing: If $y= \left(\frac{e^x- 1}{x}\right)^{1/x}$ then $ln(y)= (ln(e^x- 1)- ln(x))/x$. Both of those go to negative infinity. Which goes to negative infinity faster?
$f(x) = \left(\frac{e^x-1}x\right)^{\frac{1}{x}}$$f(x) = \left(\frac{e^x-1}x\right)^{\frac{1}{x}}$ – Kenny Lau Apr 04 '16 at 14:47