I know that the limit as n goes to infinity of ((n-2)/n)^n=e^-2 but I do not know the steps to arrive there. Can someone help me?

Also how do you enter math so that it actually looks like math and not the "gobbelty gook" above

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- Feb 8th 2010, 01:12 PMSlyprinceCan Someone help me take this limit?
I know that the limit as n goes to infinity of ((n-2)/n)^n=e^-2 but I do not know the steps to arrive there. Can someone help me?

Also how do you enter math so that it actually looks like math and not the "gobbelty gook" above - Feb 8th 2010, 01:24 PMtonio

Do you already know that $\displaystyle \lim_{n\to\infty}\left(1+\frac{1}{f(n)}\right)^{f( n)}=e $ for any function $\displaystyle f(n)\,\,\,s.t.\,\,\, f(n)\xrightarrow [n\to\infty]{} \pm\infty$ ? If you do then:

$\displaystyle \left(\frac{n-2}{n}\right)^n=\left[\left(1+\frac{1}{-n\slash 2}\right)^{-n\slash 2}\right]^{-2}\xrightarrow [n\to\infty]{}e^{-2}$

Tonio - Feb 8th 2010, 01:24 PMicemanfan
You need to learn how to use LaTeX. There is a subforum on this website for learning how to use it.

- Feb 8th 2010, 01:29 PMHallsofIvy
To answer your second question, use "LaTex" by staring with "[math ]" and ending with "[/math ]" without the last space.

For example, $\displaystyle \lim_{n\to\infty}\left(\frac{n- 2}{n}\right)^n$.

Click on that to see the code. There is also a "LaTex tutorial" on this forum.

To answer your first question, rewrite the formula as $\displaystyle \left(1- \frac{2}{n}\right)^n$ and let $\displaystyle m= \frac{n}{2}$ so n= 2m. Now it becomes $\displaystyle \left(1- \frac{1}{m}\right)^{2m}$$\displaystyle = \left[\left(1- \frac{1}{m}\right)^m\right]^2$.

Before continuing, I would need to know- what definition of "e" are you using?

Once again, Tonio got in just ahead of me! - Feb 8th 2010, 02:11 PMSlyprince
where e= the limit of (1+1/n) as n goes to infinity.

Thanks Tonio but it seems like you went through a few steps which I did not follow. Can someone fill in the gaps. I'm in the second calculus in college (series and multivariable calculus) and I just recently learned that that was the definition of e. I was never taught that so now I'm lost when it comes to these kind of limits.

Do you guys get paid for this. This seems a little too convenient. I was recently online and I tried to get tutored for math online and the fee was outrageous (something like $30 a month) and then I recently stumbled on this website. I guess what I'm trying to say is do you guys do this out of the goodness in your heart or do you get compensated? - Feb 8th 2010, 06:38 PMtonio