# Limit Problem

• Feb 25th 2010, 08:01 PM
crymorenoobs
Limit Problem
Hi, I'm trying to work out (limit as n=>infinity) (n * (sqrt(1 + 1/n) - 1)).

I know that by letting n = 1/t and then using l'hopitals this equals 1/2.
However, I need to use only basic limit rules (add/sub/mult/div and composition).

I feel like I am really close to the answer because I have been able to rewrite this as (n * (1 + 1/n)^(1/2)) - n.

I noted that:
ln(n * (1 + 1/n)^(1/2))) - ln(n)
= ln((1 + 1/n)^(1/2))
= 1/2(ln(1+1/n))
and since lim(1+1/n) = e as n=>inf
= 1/2(ln(e))
= 1/2

I cannot figure out any way to actually bring the ln into the equation like that though...any thoughts? Suggestions would be much appreciated.
• Feb 25th 2010, 11:02 PM
Hello crymorenoobs

Welcome to Math Help Forum!
Quote:

Originally Posted by crymorenoobs
Hi, I'm trying to work out (limit as n=>infinity) (n * (sqrt(1 + 1/n) - 1)).

I know that by letting n = 1/t and then using l'hopitals this equals 1/2.
However, I need to use only basic limit rules (add/sub/mult/div and composition).

I feel like I am really close to the answer because I have been able to rewrite this as (n * (1 + 1/n)^(1/2)) - n.

I noted that:
ln(n * (1 + 1/n)^(1/2))) - ln(n)
= ln((1 + 1/n)^(1/2))
= 1/2(ln(1+1/n))
and since lim(1+1/n) = e as n=>inf
= 1/2(ln(e))
= 1/2

I cannot figure out any way to actually bring the ln into the equation like that though...any thoughts? Suggestions would be much appreciated.

What about the Binomial Expansion?
$\left(1+\frac1n\right)^{\frac12}$
$=1+\left(\frac12\right)\left(\frac1n\right)+\frac{ 1}{2!}\left(\frac12\right)\left(-\frac12\right)\left(\frac1n\right)^2+\frac{1}{3!}\ left(\frac12\right)\left(-\frac12\right)\left(-\frac32\right)\left(\frac1n\right)^3+...$
Subtract $1$, multiply by $n$, and you're there.

• Feb 25th 2010, 11:09 PM
crymorenoobs
Oohhhhhh, that would do it. Thanks so much.
• Feb 26th 2010, 04:31 AM
HallsofIvy
Unfortunately, ln(a- b) is NOT equal to ln(a)- ln(b) so your first sentence using ln is incorrect.
• Feb 26th 2010, 06:38 AM
crymorenoobs
Yeah, I was wondering if there was maybe a trick I didn't know about or something because that was the only way I could see to incorporate 1/2..by bringing down the exponent. Guess not though :P
• Feb 26th 2010, 07:16 AM
skeeter
$
\lim_{n \to \infty} n\left[\sqrt{1+\frac{1}{n}} -1\right] \cdot \frac{\sqrt{1+\frac{1}{n}} +1}{\sqrt{1+\frac{1}{n}}+1}
$

$
\lim_{n \to \infty} n \cdot \frac{\left(1+\frac{1}{n}\right) - 1}{\sqrt{1+\frac{1}{n}}+1}
$

$
\lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{2}
$