# interesting limit

• Aug 29th 2007, 06:38 PM
putnam120
interesting limit
For all of the following prove without using l'Hopital's rule.

i) Does $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{\sq rt{n}}\right)^n$ converge? If so what is its limit? If not prove that it diverges.

ii) For what values $p\in\mathbb{R}$ does $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n^p }\right)^n$ converge? If possible give a formula for the converging value.
• Aug 29th 2007, 06:59 PM
ThePerfectHacker
Quote:

Originally Posted by putnam120
For all of the following prove without using l'Hopital's rule.

i) Does $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{\sq rt{n}}\right)^n$ converge? If so what is its limit? If not prove that it diverges.

ii) For what values $p\in\mathbb{R}$ does $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n^p }\right)^n$ converge? If possible give a formula for the converging value.

Maybe this will help.
• Aug 29th 2007, 08:42 PM
putnam120
thanks. I solved them by using the binomial theorem but that way works also. thanks once more.
• Aug 29th 2007, 08:58 PM
CaptainBlack
Quote:

Originally Posted by putnam120
For all of the following prove without using l'Hopital's rule.

i) Does $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{\sq rt{n}}\right)^n$ converge? If so what is its limit? If not prove that it diverges.

ii) For what values $p\in\mathbb{R}$ does $\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n^p }\right)^n$ converge? If possible give a formula for the converging value.

You need to know that:

$
\lim_{x \to \infty}\left(1+\frac{1}{x}\right)^x = e
$

Then for $x$ large enough:

$
\left(1+\frac{1}{x}\right)^x=e+b_n
$

and $|b_n|<1$.

So:

$
(e-1)^{\sqrt{n}}<\left(1+\frac{1}{\sqrt{n}}\right)^n< (e+1)^{\sqrt{n}}
$

and the rest of part (i)should follow from the squeeze theorem.

RonL