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

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

$$y=\lim _{ n\rightarrow \infty }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n^{ 2 } } } \frac { 1 }{ { e }^{ n } } \\ z=\lim _{ n\rightarrow \infty }{ { \left( 1-\frac { 1 }{ n } \right) }^{ n^{ 2 } } } { e }^{ n }\\ yz\quad =\quad \frac { 1 }{ e } $$

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.

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 $\frac{1}{\sqrt{e}}$.