Limit

Nils-Ake · 2008-10-30 08:02:54

Hi! Could you help me calculate this limit?

mathsyperson · 2008-10-30 08:21:03

As n gets large, the base is approximated by 1 + 2/n and the exponent is approximated by n.

So the limit is

Nils-Ake · 2008-10-30 09:15:51

Thanks mathsyperson. I don't understand why the base is approximated by 1 + 2/n. Could you please explain me that part?

mathsyperson · 2008-10-30 10:44:39

You can rewrite the numerator as n² - n + 1 + 2n.
Then the first three terms equal the denominator and you can split them off to get

1 + 2n/(n² - n + 1).

As n grows, 1 << n << n², and so the denominator of that new fraction is approximated by n². Therefore, you have 1 + 2n/n² = 1 + 2/n.

(To be honest, I wouldn't be confident about doing that usually. I cheated and evaluated the function for large values of n. It converged to e², and I made the above post with that knowledge.)

Nils-Ake · 2008-10-30 11:05:13

You can rewrite the numerator as n² - n + 1 + 2n

Oh I missed that, I understand it now! Thank you

Nils-Ake · 2008-10-30 11:14:53

mathsyperson wrote:

(To be honest, I wouldn't be confident about doing that usually. I cheated and evaluated the function for large values of n. It converged to e², and I made the above post with that knowledge.)

I don't know if this is what you mean, but I tried this in maple: seq(limit((1+i/n)^n, n=infinity), i=1..10);

It outputs e^i.

mathsyperson · 2008-10-30 12:12:48

Sorry, I meant I wasn't sure if those lower-order terms I neglected were allowed to be neglected.
It certainly doesn't always work: (1+1/n)^n and 1^n converge to different things.

I knew my limit was e², but I wouldn't have known whether it was the same as yours.

Math Is Fun Forum

#1 2008-10-30 08:02:54

Limit

#2 2008-10-30 08:21:03

Re: Limit

#3 2008-10-30 09:15:51

Re: Limit

#4 2008-10-30 10:44:39

Re: Limit

#5 2008-10-30 11:05:13

Re: Limit

#6 2008-10-30 11:14:53

Re: Limit

#7 2008-10-30 12:12:48

Re: Limit

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