Evaluating $\lim_{n\rightarrow\infty} (\frac{(1+\frac{1}{n})^n}{e})^n$ We know that $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$ and so I thought the approach to evaluating the limit in the question would be to just use this fact and substitute it into the numerator. This approach would tell us the above limit evaluates to $1$. However, that does not seem to be the correct limiting value. In fact, it evaluates to $\frac {1}{\sqrt{e}}$. Why is this so?
 A: This is a companion to Eevee Trainer's answer: The same (il)logic that says 
$$\left((1+1/n)^n\over e\right)^n\to\left(e\over e\right)^n=1^n=1$$ 
would also say 
$$\left(1+{1\over n}\right)^n\to(1+0)^n=1^n=1$$
A: Let $x=\frac1n$. Using
$$ \ln(1+x)=x-\frac12x^2+O(x^3)$$
one has
\begin{eqnarray}
&&\lim_{n\rightarrow\infty} \ln\bigg(\frac{(1+\frac{1}{n})^n}{e}\bigg)^n\\
&=&\lim_{n\rightarrow\infty} n\bigg(n\ln(1+\frac{1}{n})-1\bigg)\\
&=&\lim_{n\rightarrow\infty} \frac{n\ln(1+\frac{1}{n})-1}{\frac1n}\\
&=&\lim_{x\rightarrow0} \frac{\frac1x\ln(1+x)-1}{x}\\
&=&\lim_{x\rightarrow0} \frac{-\frac12x+O(x^2)}{x}\\
&=&-\frac12.
\end{eqnarray}
So
$$ \lim_{n\rightarrow\infty} \bigg(\frac{(1+\frac{1}{n})^n}{e}\bigg)^n=e^{-1/2}. $$
A: Since everyone else has decided to cover ways to calculate the limit (despite a comment from you mentioning that you want to know why you're wrong and not how to solve the exercise), I'll answer focusing on that. First, simplifying, you have
$$\lim_{n \to \infty} \frac{ ((1+1/n)^n)^n }{e^n}$$
In replacing the top expression with $e^n$, you implicitly assume that you can take the limit inside as so, with your substitution in blue:
$$\lim_{n \to \infty} \left( \left( 1 + \frac 1 n \right)^n \right)^n =  \left(\color{blue}{\lim_{n \to \infty} \left( 1 + \frac 1 n \right)^n} \right)^n =\color{blue}{e}^n$$
However, you have a dependence on $n$ on the outer parentheses, and thus this step is not justified. You can only move a limit inside a (continuous) function when you're not suddenly moving a dependence on $n$ to the outside.
A: Just for your curiosity.
We can have much more than the limit itself composing Taylor series
$$a_n=\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right)^n\implies \log(a_n)=n^2 \log\left(1+\frac{1}{n}\right)-n$$Now, using Taylor expansion
$$\log(a_n)=n^2\left(\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{3 n^3}-\frac{1}{4 n^4}+O\left(\frac{1}{n^5}\right)\right)-n$$
$$\log(a_n)=-\frac{1}{2}+\frac{1}{3 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
$$a_n=e^{\log(a_n)}=\frac 1 {\sqrt e}\left(1+\frac{1}{3 n}-\frac{7}{36 n^2}\right)+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
Moreover, this gives you a shortcut method for a quick evaluation of $a_n$ even for small values of $n$. For example, the exact calculation of $a_{10}= 0.625639$ while the above truncated expansion gives $\frac{3713}{3600 \sqrt{e}}=0.625569$.
A: Rewrite the function as $\phi(n) = e^{\log \phi(n)} = e^{n^2 \log (1+\frac{1}{n}) - n} = e^{n^2(\frac{1}{n} -\frac{1}{2n^2} + O(\frac{1}{n^3}))- n} = e^{-\frac{1}{2} + O(\frac{1}{n})} = e^{-\frac{1}{2}}$
I used Maclaurin series expansion here
A: Since $ 1^{\infty} $ is an indeterminate form, we can not just say that "since $ \lim\limits_{n\to +\infty}{\left(1+\frac{1}{n}\right)^{n}}=\mathrm{e} $, the limit of the ratio raised to the $ n^{\mathrm{th}} $ power will be just $ 1 $".
Let's proceed without using either L'Hospital's rule or series expansion :
Let $ n $ be a positive integer, we have :
\begin{aligned}\lim_{n\to +\infty}{\left(\frac{\left(1+\frac{1}{n}\right)^{n}}{\mathrm{e}}\right)^{n}}&=\lim_{n\to +\infty}{\mathrm{e}^{-n}\left(1+\frac{1}{n}\right)^{n^{2}}}\\ &=\lim_{n\to +\infty}{\exp{\left(-n+n^{2}\ln{\left(1+\frac{1}{n}\right)}\right)}}\\ &=\lim_{n\to +\infty}{\exp{\left(-\frac{\frac{1}{n}-\ln{\left(1+\frac{1}{n}\right)}}{\frac{1}{n^{2}}}\right)}}\end{aligned}
Now let $ x\geq 0 $, note that : \begin{aligned} \frac{x-\ln{\left(1+x\right)}}{x^{2}}&=\int_{0}^{1}{\frac{1-y}{\left(1+xy\right)^{2}}\,\mathrm{d}y}\\ &=\int_{0}^{1}{\left(1-y\right)\mathrm{d}y}-\int_{0}^{1}{\left(1-y\right)\left(1-\frac{1}{\left(1+xy\right)^{2}}\right)\mathrm{d}y}\\ &=\frac{1}{2}-x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y} \end{aligned}
Since $ x\int_{0}^{1}{\frac{y\left(2+xy\right)}{\left(1+xy\right)^{2}}\,\mathrm{d}y}\leq x\left(2+x\right)\int_{0}^{1}{y\,\mathrm{d}y}=\frac{x\left(2+x\right)}{2}\underset{x\to 0}{\longrightarrow}0 $, we get that : $$ \lim_{x\to 0}{\frac{x-\ln{\left(1+x\right)}}{x^{2}}}=\frac{1}{2} $$
Hence : $$ \lim_{n\to +\infty}{\frac{\frac{1}{n}-\ln{\left(1+\frac{1}{n}\right)}}{\frac{1}{n^{2}}}}=\frac{1}{2} $$
Which means : $$ \lim_{n\to +\infty}{\left(\frac{\left(1+\frac{1}{n}\right)^{n}}{\mathrm{e}}\right)^{n}}=\lim_{n\to +\infty}{\exp{\left(-\frac{\frac{1}{n}-\ln{\left(1+\frac{1}{n}\right)}}{\frac{1}{n^{2}}}\right)}}=\exp{\left(-\frac{1}{2}\right)} $$
A: $1^\infty$ is an indeterminate form.
Taking logarithm, we get
$$n\Bigl(\ln((1+\frac 1n)^n)-1\Bigr)=$$
$$n\Bigl(n\ln(1+\frac 1n)-1\Bigr)=$$
$$n\Bigl(n(\frac 1n-\frac{1}{2n^2}+\frac{1}{n^2}\epsilon(n))-1\Bigr)=$$
$$\frac{-1}{2}+\epsilon(n)$$
the limit is $$\sqrt{\frac 1e}$$
