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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.

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    do you know the technique of taking logs on both sides? – 2017-01-27
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    $\lim_{x\rightarrow0}\Bigg[\frac{(1+x)^{1/x}}{e}\Bigg]^{1/x}=\lim_{x\rightarrow0}\frac{(1+x)^{1/x^2}}{e^{1/x}}=\lim_{x\rightarrow0}\Bigg[\frac{(1+x)^{1/x^2}}{\lim_{x\rightarrow0}(1+x)^{1/x}}\Bigg]=\lim_{x\rightarrow0}\Bigg[\frac{(1+x)^{1/x^2}}{(1+x)^{1/x}}\Bigg]\\=\lim_{x\rightarrow0}(1+x)^{1/x^2-1/x}=\lim_{x\rightarrow0}(1+x)^{\frac{1-x}{x^2}}=\lim_{x\rightarrow0}(1+x)^{\frac{1-x}{x^2}}$ Can we think something like this??? – 2017-01-27

4 Answers 4

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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 $Y = e^{\frac{-1}2}$.

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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 $x \to 0$, $$ \left[\frac{(1+x)^{1/x}}{e}\right]^{1/x}=\left[\frac{e^{\large 1-\frac{x}2+O(x^2)}}{e}\right]^{1/x}=e^{-1/2+O(x)}. $$

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    Well, I forgot to add that I don't know anything about the Taylor series. Thanks for the effort. – 2017-01-28
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Using L'Hospital rule twice we get $$\lim _{ x\rightarrow 0 }{ { \left[ \frac { { \left( 1+x \right) }^{ \frac { 1 }{ x } } }{ e } \right] }^{ \frac { 1 }{ x } } } ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \ln { \left[ \frac { { \left( 1+x \right) }^{ \frac { 1 }{ x } } }{ e } \right] } } }=e^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left[ \frac { { \left( 1+x \right) }^{ \frac { 1 }{ x } } }{ e } \right] } }{ x } } }=e^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 1 }{ x } \ln { \left( 1+x \right) -1 } }{ x } } }=\\ =e^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+x \right) -x } }{ { x }^{ 2 } } } }\overset { L'\quad Hospital }{ = } e^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 1 }{ x+1 } -1 }{ 2x } } }\overset { L'\quad Hospital }{ = } e^{ \lim _{ x\rightarrow 0 }{ \frac { -\frac { 1 }{ { \left( x+1 \right) }^{ 2 } } }{ 2 } } }={ e }^{ -1/2 }\\ \\ \\ $$

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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 $[\frac{(1+x)^{\frac{1}{x}}}{e}]^{1/x}=e^{1/x\ln(e^{\frac{1}{x}\ln(1+x)-1})}=e^{\frac{1}{x^2}ln(1+x)-1}$