Evaluate of $\lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n}$ 
Evaluate the limit
$$
\lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n}
$$

My Attempt:
$$
\lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n}
$$
How can I solve the problem from this point?
 A: This answer is morally the same thing as what has already been posted, but I have a personal preference to avoid using $o$-notation on any problem I think can be done without it.  It's elegant and quick, but I expect students to have more trouble with it than not when I would expect a problem like this to arise.  It is often not even introduced to them.  So I will use L'Hospital.
As in the other approaches, we take the logarithm of the expression, and exponentiate the limit once we've found it to get the final answer.
First, we get it into a form we can use L'Hospital on:
$$
\begin{align}
n^2\ln\left( 1+\frac{1}{n}\right) - n &= \frac{\ln(1+1/n)}{1/n^2} - \frac{1/n}{1/n^2}\\
&= \frac{\ln(1+1/n)-1/n}{1/n^2}
\end{align}
$$
It's easy to verify that L'Hospital applies here when $n\rightarrow\infty$.
Now,
$$
\begin{align}
\lim_{n\rightarrow\infty} \frac{\ln(1+1/n)-1/n}{1/n^2} &\overset{L'H}{=}\lim\limits_{n\rightarrow\infty}\frac{ \frac{1}{1+1/n}\cdot\frac{-1}{n^2}+\frac{1}{n^2}}{-2/n^3}\\
&=\lim_{n\rightarrow\infty}\frac{1}{2}\left( \frac{n}{1+1/n}-n\right)\\
&= \lim_{n\rightarrow\infty}-\frac{1}{2}\left(\frac{1}{1+1/n}\right)\\
&=-\frac{1}{2}
\end{align}
$$
So the answer is $e^{-1/2}=\frac{1}{\sqrt{e}}$.
A: $$y=\lim _{ n\rightarrow \infty  }{ { \left( 1+\frac { 1 }{ n }  \right)  }^{ n^{ 2 } } } \frac { 1 }{ { e }^{ n } } \\ z=\lim _{ n\rightarrow \infty  }{ { \left( 1-\frac { 1 }{ n }  \right)  }^{ n^{ 2 } } } { e }^{ n }\\ yz\quad =\quad \frac { 1 }{ e } $$
If you can show $y=z$ then you have a simple proof (I tried hard but I couldn't). Wolframalpha says yes, they are equal.
A: For the logarithm of the expression, we have $$n^2\log\left(1+\frac{1}{n}\right)-n \in n^2\left(\frac{1}{n}-\frac{1}{2n^2}+o(1/n^2)\right)-n \in -\frac{1}{2}+o(1).$$ Hence, our function is a member of $e^{-\frac{1}{2}+o(1)}$, which implies that the limit is $\frac{1}{\sqrt{e}}$.
A: A proof without Taylor, L'Hopital or the $o()$ business: For $h>0,$
$$\tag 1 \ln (1+h) = \int_1^{1+h}\frac{dt}{t} = \int_0^{h}\frac{dt}{1+t}.$$
Now for $t>0,1-t \le 1/(1+t) \le 1-t+t^2.$ This is simple to verify. So using $(1)$ we get
$$\tag 2 h - h^2/2  \le \ln(1+h) \le h - h^2/2 +h^3/3, h> 0.$$
So in the given problem we apply $\ln$ to the expression and then use $(2)$ with $h=1/n.$ The limit of $-1/2$ falls right out. So the limit of the original expression is $e^{-1/2}.$
A: You have 
$$ \log(1 + x) = x - x^2/2 +O(x^3).$$
so
$$\log(1 + 1/n) = {1\over n} - {1\over 2n^2} + O(1/n^3).$$
Threfore
$$ n^2 \log(1 + 1/n) = n - 1/2 + O(1/n).$$
Your limit converges to ${1\over \sqrt{e}} \approx .606531.$
A: Consider $$\lim_{n\rightarrow \infty}({n+1\over n})^{n^2}\cdot{1\over e^n}.$$ This limit has an indeterminate form. Let $y=(1+{1\over n})^{n^2}\cdot {1\over e^n}$. Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain $\ln(y)=n^2\ln(1+{1\over n})-n$. We can rewrite this as $\ln(y)={{\ln(1+{1\over n})-{1\over n}}\over {1\over n^2}}$. The $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}{{\ln(1+{1\over n})-{1\over n}}\over {1\over n^2}}$$ which has the indeterminate form $0\over 0$. Using L'Hospital's Rule we see that $$\lim_{n\rightarrow \infty}\ln(y)=\lim_{n\rightarrow \infty}{{1\over (1+{1\over n})}\cdot {-1\over n^2}+{1\over n^2}\over {-2\over n^3}}.$$ Cancelling the ${-1\over n^2}$ on the numerator and the denominator this simplifies to $$\lim_{n\rightarrow \infty} \ln(y)=\lim_{n\rightarrow \infty}{{1\over 1+{1\over n}}-1\over {2\over n}}=\lim_{n\rightarrow\infty}{{-1\over 2(1+{1\over n})}}={-1\over 2}.$$ So far we have computed the limit of $\ln(y)$, what we really want is the limit of $y$. We know that $y=e^{\ln(y)}$. So $$\lim_{n\rightarrow \infty}({n+1\over n})^{n^2}\cdot{1\over e^n}=\lim_{n\rightarrow\infty}y=\lim_{n\rightarrow\infty}e^{\ln(y)}=e^{-1\over 2}={1\over \sqrt e}.$$ Thus $$\lim_{n\rightarrow \infty}({n+1\over n})^{n^2}\cdot{1\over e^n}={1\over \sqrt e}.$$
A: Setting $h=\frac{1}{n}$
$$a_n =\left(1+\dfrac{1}{n}\right)^{n^2} =\exp\left(\frac{\ln(1+\frac{1}{n})}{\frac{1}{n^2}}\right)=\exp\left(\frac{\ln(1+h)}{h^2}\right)$$
Thus,
$$\lim_{n\to +\infty} a_ne^{-n} =\lim_{h\to 0} = \exp\left(-\frac{1}{h}+\frac{\ln(1+h)}{h^2}\right)=e^{-\frac{1}{2}}$$
Given, that we know by Schwartz  derivative that, if a function is $C^2$ near $x = 0$ we have,
$$\color{red}{\frac{f''(0)}{2} =\lim_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}}$$
taking $f(x) = \ln(1+x)$, $f(0)= 0$, $f'(0) =1$,$f''(0) =-1$
$$\color{red}{-\frac{1}{2} =\lim_{h\to 0} \frac{\frac{ \ln(1+h) }{h}-1}{h} =\lim_{h\to 0} -\frac{1 }{h}+\frac{ \ln(1+h) }{h^2}}$$
See here: How to prove Schwarz derivative $\frac{f''(0)}{2} =\lim_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}$ without Taylor expansion or L'Hopital rule?
