Finding limit of $\lim_{x\to 0^+}⁡\{[(1+x)^{1/x}]/e\}^{1/x}$ This is a question given in our weekly test.
$$f = \lim_{x\to 0^+}⁡\{[(1+x)^{1/x}]/e\}^{1/x}.$$
Find the value of $f$. I tried to use 1^ infinity form but I didn't get it. So anybody please help me.
 A: Your limit $f$ exists if and only if its logarithm exists:
\begin{align}
\log f
&=\lim_{x\to0^+}\log\bigl(\bigl((1+x)^{1/x})/e\bigr)^{1/x}\bigr)
\\[6px]
&=\lim_{x\to0^+}\frac{\dfrac{1}{x}\log(1+x)-1}{x}
\\[6px]
&=\lim_{x\to0^+}\frac{\log(1+x)-x}{x^2}
\\[6px]
&=\lim_{x\to0^+}\frac{x-x^2/2+o(x^2)-x}{x^2}=-\frac{1}{2}
\end{align}
If you don't trust Taylor expansions (or cannot use them),
$$
\lim_{x\to0^+}\frac{\log(1+x)-x}{x^2}
\overset{*}{=}
\lim_{x\to0^+}\frac{\dfrac{1}{1+x}-1}{2x}
=
\lim_{x\to0^+}\frac{-x}{2x(1+x)}
$$
(where $\overset{*}{=}$ denotes an application of l'Hôpital).
A: \begin{align}
f&=\lim_{x\to0}\left(\frac{(1+x)^{1/x}}e\right)^{1/x}\\
&=\lim_{x\to0}\left(\frac1{e^x}+\frac{x}{e^x}\right)^{1/x^2}\\
&=\lim_{x\to0}\left(1+\frac{1+x-e^x}{e^x}\right)^{1/x^2}\\
&=\lim_{x\to0}\left(\left(1+\frac{1+x-e^x}{e^x}\right)^{\cfrac{e^x}{1+x-e^x}}\right)^{\cfrac{1+x-e^x}{x^2e^x}}\\
&=\lim_{x\to0}\quad e^{\cfrac{1+x-e^x}{x^2}\cfrac1{e^x}}\\
&=e^{\ \lim_{x\to0}\cfrac{1+x-e^x}{x^2}\cfrac1{e^x}}\\
&=e^{\ \lim_{x\to0}\cfrac{1+x-e^x}{x^2}}\\
&\overset{\text{L'Hopital}}{=}e^{\ \lim_{x\to0}\cfrac{1-e^x}{2x}}\\
&\overset{\text{L'Hopital}}{=}e^{\ \lim_{x\to0}\cfrac{-e^x}{2}}\\
&=e^{-1/2}
\end{align}
A: $$\lim\limits_{x\to 0^{+}}\left(\frac{(1+x)^{1/x}}{x}\right)^{1/x}=\lim\limits_{x\to 0^{+}}e^{-1/x}\left((x+1)^{1/x}\right)^{1/x}=\lim\limits_{x\to 0^{+}}\exp\left(\frac{\ln((x+1)^{1/x})}{x}-\frac 1 x\right)$$
$$=\exp\left(\lim\limits_{x\to 0^{+}}\left(\frac{\ln\left((1+x)^{1/x}\right)}{x}-\frac 1 x\right)\right)^{1/x}=\exp\left(\lim\limits_{x\to 0^{+}}\left(\frac{\frac{\ln(x+1)}{x}}{x}-\frac 1 x\right)\right)\\
=\exp\left(\lim\limits_{x\to 0^{+}}\frac{\ln(x+1)-x}{x^2}\right)$$
$$\overset{\text{L'Hopital}}{=}\exp \left(\lim\limits_{x\to 0^{+}}\underbrace{-\frac{1}{2(x+1)}}_{\to -1/2}\right)=e^{-1/2}$$
