Limits

Printable View

November 29th 2011, 04:23 AM
Abhi

Limits

Lim x tends to infinity ( e / ( 1 + 1 / x )^x ) ^x
Pls help me...
November 29th 2011, 05:35 AM
Prove It

Re: Limits

Quote:

Originally Posted by Abhi

Lim x tends to infinity ( e / ( 1 + 1 / x )^x ) ^x
Pls help me...

Is this $\displaystyle \begin{align*} \lim_{x \to \infty}\left[\frac{e}{\left(1 + \frac{1}{x}\right)^x}\right]^x \end{align*}$ ?
November 29th 2011, 05:40 AM
Abhi

Re: Limits

Quote:

Originally Posted by Prove It

Is this $\displaystyle \begin{align*} \lim_{x \to \infty}\left[\frac{e}{\left(1 + \frac{1}{x}\right)^x}\right]^x \end{align*}$ ?

yes
November 29th 2011, 05:51 AM
sbhatnagar

Re: Limits

Hello Friends!

Quote:

Solve $\lim_{x \to \infty}[\frac{e}{(1+\frac{1}{x})^x}]^x$

$\begin{align*} \lim_{x \to \infty}[\frac{e}{(1+\frac{1}{x})^x}]^x &= \exp \lim_{x \to \infty} x\ln[\frac{e}{(1+\frac{1}{x})^x}] \\ &=\exp \lim_{x \to \infty}x[\ln{e}-x\ln{(1+\frac{1}{x}})] \\ &=\exp\lim_{x \to \infty} [x-x^2\ln{(1+\frac{1}{x}})] \end{align*}$

Quote:

For the limit $\lim_{x \to \infty} [x-x^2\ln{(1+\frac{1}{x}})]$

$\begin{align*} \lim_{x \to \infty} [x-x^2\ln{(1+\frac{1}{x}})] &= \lim_{x \to \infty} [x-x^2\left \{ \frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-... \right \}] \\ &=\lim_{x \to \infty}[x-x+\frac{1}{2}-\frac{1}{3x}+...] \\ &=\frac{1}{2} \end{align*}$

Therefore: $\lim_{x \to \infty}[\frac{e}{(1+\frac{1}{x})^x}]^x=\exp\lim_{x \to \infty} [x-x^2\ln{(1+\frac{1}{x}})]=\exp(\frac{1}{2})=\sqrt{e}$

All times are GMT -8. The time now is 02:58 AM.

Copyright © 2005-2015 Math Help Forum. All rights reserved.

Copyright © 2005-2014 Math Help Forum. All rights reserved.