# Thread: Limit Problem

1. ## 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.

2. Hello crymorenoobs

Welcome to Math Help Forum!
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.

3. Oohhhhhh, that would do it. Thanks so much.

4. Unfortunately, ln(a- b) is NOT equal to ln(a)- ln(b) so your first sentence using ln is incorrect.

5. 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

6. $
\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}
$