poveli1e

Answered

2022-01-23

Evaluate the limit

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

Answer & Explanation

Addisyn Thompson

Expert

2022-01-24Added 16 answers

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

egowaffle26ic

Expert

2022-01-25Added 7 answers

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$y={(1+\frac{1}{n})}^{{n}^{2}}\cdot \frac{1}{{e}^{n}}$ . Taking the natural logarithm of both sides of the equation and simplifying using the rules of logarithms we obtain $\mathrm{ln}\left(y\right)={n}^{2}\mathrm{ln}(1+\frac{1}{n})-n$ . We can rewrite this as $\mathrm{ln}\left(y\right)=\frac{\mathrm{ln}(1+\frac{1}{n})-\frac{1}{n}}{\frac{1}{{n}^{2}}}$

The

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{ln}\left(y\right)=\underset{n\to \mathrm{\infty}}{lim}\frac{\mathrm{ln}(1+\frac{1}{n})-\frac{1}{n}}{\frac{1}{{n}^{2}}}$

which has the indeterminate form$\frac{0}{0}$ . Using LHospitals Rule we see that

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{ln}\left(y\right)=\underset{n\to \mathrm{\infty}}{lim}\frac{\frac{1}{(1+\frac{1}{n})\cdot \frac{-1}{{n}^{2}}+\frac{1}{{n}^{2}}}}{\frac{-2}{{n}^{3}}}$

Canceling the$\frac{-1}{{n}^{2}}$ on the numerator and the denominator this simplifies to

$\underset{n\to \mathrm{\infty}}{lim}\mathrm{ln}\left(y\right)=\underset{n\to \mathrm{\infty}}{lim}\frac{\frac{1}{1+\frac{1}{n}}-1}{\frac{2}{n}}=\underset{n\to \mathrm{\infty}}{lim}\frac{-1}{2(1+\frac{1}{n})}=\frac{-1}{2}$

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

This limit has an indeterminate form. Let

The

which has the indeterminate form

Canceling the

So far we have computed the limit of

Thus

RizerMix

Expert

2022-01-27Added 437 answers

Setting $h=\frac{1}{n}$ ${a}_{n}=(1+\frac{1}{n}{)}^{{n}^{2}}=exp(\frac{\mathrm{ln}(1+\frac{1}{n})}{\frac{1}{{n}^{2}}})=exp(\frac{\mathrm{ln}(1+h)}{{h}^{2}})$ Thus,$\underset{n\to +\mathrm{\infty}}{lim}{a}_{n}{e}^{-n}=\underset{h\to 0}{lim}exp(-\frac{1}{h}+\frac{\mathrm{ln}(1+h)}{{h}^{2}})={e}^{-\frac{1}{2}}$ Given, that we know by Schwartz derivative that, if a function is ${C}^{2}$ near $x=0$ we have,$\frac{{f}^{\u2033}(0)}{2}=\underset{x\to 0}{lim}\frac{\frac{f(x)-f(0)}{x}-{f}^{\prime}(0)}{x}$ taking $f(x)=\mathrm{ln}(1+x),\text{}f(0)=0,\text{}{f}^{\prime}(0)=1,\text{}{f}^{\u2033}(0)=-1$ $-\frac{1}{2}=\underset{h\to 0}{lim}\frac{\frac{\mathrm{ln}(1+h)}{h}-1}{h}=\underset{h\to 0}{lim}-\frac{1}{h}+\frac{\mathrm{ln}(1+h)}{{h}^{2}}$

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