Finding $ \lim_{x \to 0}\left[\frac{(1+x)^{1/x}}{e}\right]^{1/x}.$ Evaluate
$$
\lim_{x \to 0}\left[\frac{(1+x)^{1/x}}{e}\right]^{1/x}
$$
The answer is $$\ e^{-1/2}$$ Any help is appreciated.
 A: Hint. By using the Taylor series expansion, as $x \to 0$, one has
$$
\log(1+x)=x-\frac{x^2}2+O(x^3)
$$ giving
$$
\begin{align}
(1+x)^{1/x}&=e^{\large\frac{\log(1+x)}x}
\\&=e^{\large\frac{x-\frac{x^2}2+O(x^3)}x}
\\&=e^{\large 1-\frac{x}2+O(x^2)}
\end{align}
$$ then, as $x \to 0$,
$$
\left[\frac{(1+x)^{1/x}}{e}\right]^{1/x}=\left[\frac{e^{\large 1-\frac{x}2+O(x^2)}}{e}\right]^{1/x}=e^{-1/2+O(x)}.
$$
A: Using L'Hospital rule twice we get
$$\lim _{ x\rightarrow 0 }{ { \left[ \frac { { \left( 1+x \right)  }^{ \frac { 1 }{ x }  } }{ e }  \right]  }^{ \frac { 1 }{ x }  } } ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \ln { \left[ \frac { { \left( 1+x \right)  }^{ \frac { 1 }{ x }  } }{ e }  \right]  }  }  }=e^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left[ \frac { { \left( 1+x \right)  }^{ \frac { 1 }{ x }  } }{ e }  \right]  }  }{ x }  }  }=e^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 1 }{ x } \ln { \left( 1+x \right) -1 }  }{ x }  }  }=\\ =e^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+x \right) -x }  }{ { x }^{ 2 } }  }  }\overset { L'\quad Hospital }{ = } e^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 1 }{ x+1 } -1 }{ 2x }  }  }\overset { L'\quad Hospital }{ = } e^{ \lim _{ x\rightarrow 0 }{ \frac { -\frac { 1 }{ { \left( x+1 \right)  }^{ 2 } }  }{ 2 }  }  }={ e }^{ -1/2 }\\ \\ \\ $$
A: I think this is start :
$(1+x)^{\frac{1}{x}}=e^{\frac{1}{x}\ln(1+x)}$
hence :
$\frac{(1+x)^{\frac{1}{x}}}{e}=e^{\frac{1}{x}\ln(1+x)-1}$
and $[\frac{(1+x)^{\frac{1}{x}}}{e}]^{1/x}=e^{1/x\ln(e^{\frac{1}{x}\ln(1+x)-1})}=e^{\frac{1}{x^2}ln(1+x)-1}$
A: we have $Y=\left\{\lim_{x \to 0}\left(\frac{(1 + x)^{1/x}}{e}\right)^{1/x}\right\}$
\begin{align}
\log Y &= \log\left\{\lim_{x \to 0}\left(\frac{(1 + x)^{1/x}}{e}\right)^{1/x}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{(1 + x)^{1/x}}{e}\right)^{1/x}\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\left(\frac{(1 + x)^{1/x}}{e}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x}\left(\log(1 + x)^{1/x} - 1\right)\notag\\
&= \lim_{x \to 0}\frac{\log (1 + x) - x}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{x - \dfrac{x^{2}}{2} + o(x^{2}) - x}{x^{2}}\notag\\
&= -\frac{1}{2}\notag
\end{align}
   Hence $Y = e^{\frac{-1}2}$.
