Evaluate the limit limn→∞(n+1n)n2⋅1en

First, we get it into a form we can use LHospital

Consider
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n+1}{n}\right)}^{n^2}\cdot \frac{1}{e^n}}$
This limit has an indeterminate form. Let ${\displaystyle y={(1+\frac{1}{n})}^{n^2}\cdot \frac{1}{e^n}}$. Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain ${\displaystyle \ln \left(y\right)=n^2\ln (1+\frac{1}{n})-n}$. We can rewrite this as ${\displaystyle \ln \left(y\right)=\frac{\ln (1+\frac{1}{n})-\frac{1}{n}}{\frac{1}{n^2}}}$
The
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\ln \left(y\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{\ln (1+\frac{1}{n})-\frac{1}{n}}{\frac{1}{n^2}}}$
which has the indeterminate form ${\displaystyle \frac{0}{0}}$. Using LHospitals Rule we see that
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\ln \left(y\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{(1+\frac{1}{n})\cdot \frac{-1}{n^2}+\frac{1}{n^2}}}{\frac{-2}{n^3}}}$
Canceling the ${\displaystyle \frac{-1}{n^2}}$ on the numerator and the denominator this simplifies to
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\ln \left(y\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{1+\frac{1}{n}}-1}{\frac{2}{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{-1}{2(1+\frac{1}{n})}=\frac{-1}{2}}$
So far we have computed the limit of ${\displaystyle \ln \left(y\right)}$, what we really want is the limit of y. We know that ${\displaystyle y=e^{\ln \left(y\right)}}$. So
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n+1}{n}\right)}^{n^2}\cdot \frac{1}{e^n}=\underset{n\to \mathrm{\infty }}{lim}y=\underset{n\to \mathrm{\infty }}{lim}{e}^{\ln \left(y\right)}=e^{\frac{-1}{2}}=\frac{1}{\sqrt{e}}}$
Thus