poveli1e

2022-01-23

Evaluate the limit
$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n+1}{n}\right)}^{{n}^{2}}\cdot \frac{1}{{e}^{n}}$

Expert

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

egowaffle26ic

Expert

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

RizerMix

Expert

Setting $h=\frac{1}{n}$${a}_{n}=\left(1+\frac{1}{n}{\right)}^{{n}^{2}}=exp\left(\frac{\mathrm{ln}\left(1+\frac{1}{n}\right)}{\frac{1}{{n}^{2}}}\right)=exp\left(\frac{\mathrm{ln}\left(1+h\right)}{{h}^{2}}\right)$Thus,$\underset{n\to +\mathrm{\infty }}{lim}{a}_{n}{e}^{-n}=\underset{h\to 0}{lim}exp\left(-\frac{1}{h}+\frac{\mathrm{ln}\left(1+h\right)}{{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,$\frac{{f}^{″}\left(0\right)}{2}=\underset{x\to 0}{lim}\frac{\frac{f\left(x\right)-f\left(0\right)}{x}-{f}^{\prime }\left(0\right)}{x}$taking $-\frac{1}{2}=\underset{h\to 0}{lim}\frac{\frac{\mathrm{ln}\left(1+h\right)}{h}-1}{h}=\underset{h\to 0}{lim}-\frac{1}{h}+\frac{\mathrm{ln}\left(1+h\right)}{{h}^{2}}$