Evaluate the limit.
![\[ \lim_{x \to 0} \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)^{\frac{1}{x}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b77fd23f5605b7a06b06a594994b3fce_l3.png "Rendered by QuickLaTeX.com")
First, we write the expression using the definition of the exponential,
![\begin{align*} \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)^{\frac{1}{x}} &= e^{ \frac{1}{x} \log \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)} \\[10pt] &= e^{\frac{1}{x} \left( \log (1+x)^{\frac{1}{x}} - \log e \right)} \\[10pt] &= e^{\frac{1}{x} \left( \frac{1}{x} \log(1+x) - 1 \right)} \\[10pt] &= e^{\frac{1}{x^2} \log(1+x) - \frac{1}{x}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4ef72956ae91e60950ced8dc7360061d_l3.png "Rendered by QuickLaTeX.com")
Now, considering the expression in the exponent and using the expansion of 
as 
(page 287 of Apostol) we have as ,
![\begin{align*} \frac{1}{x^2} \log(1+x) - \frac{1}{x} &= \frac{1}{x^2} \left( x - \frac{x^2}{2} + o(x^2) \right) - \frac{1}{x} \\[9pt] &= \frac{1}{x} - \frac{1}{2} + \frac{o(x^2)}{x^2} - \frac{1}{x} \\[9pt] &= -\frac{1}{2} + \frac{o(x^2)}{x^2}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b9bb4fed9e2e8ff04008bd436166a673_l3.png "Rendered by QuickLaTeX.com")
Therefore, we have
![\begin{align*} \lim_{x \to 0} \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)^{\frac{1}{x}} &= \lim_{x \to 0} e^{\frac{1}{x^2} \log(1+x) - \frac{1}{x}} \\[9pt] &= \lim_{x \to 0} e^{-\frac{1}{2} + \frac{o(x^2)}{x^2}} \\[9pt] &= e^{-\frac{1}{2}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0a714479a29fc30e5053fed677a1e705_l3.png "Rendered by QuickLaTeX.com")
Evaluate the limit.
First, we write the expression using the definition of the exponential,
Now, considering the expression in the exponent and using the expansion of as (page 287 of Apostol) we have as ,
Therefore, we have