Find the limit, as n -> infty, of n*[n - sqrt(n^2 - 4)]

[green_tea]

I need some help cause I just don't get the limes thing... I have to find:

lim n(n-sqrt(n^2-4))
n-->?

I've been trying to solve it like this:

n(n - sqrt(n^2 - 4) = n^2 - n * sqrt(n^2-4) = n^2 - n*n*sqrt(1 - 4/n^2)= n^2 * (1- sqrt(1 - 4/n^2))

and I thought that since 4/n^2 ---> 0 when n--->? the whole thing would go towards ? * (1-sqrt(1-0)) = ? * (1-1) = ? * 0 = 0 cause any number times 0 is 0 ,right? So ? * 0 should be 0? Or am I wrong?

Anyway, the answer is 2 ... :? So I suppose I did something wrong? Is it the ? * 0 thing? And how did they get 2???

 

[skeeter]

Re: limes n-->? help!

\(\displaystyle n(n - \sqrt{n^2 - 4}) \cdot \frac{n + \sqrt{n^2 - 4}}{n + \sqrt{n^2 - 4}} =\)

\(\displaystyle \frac{n[n^2 - (n^2 - 4)]}{n + \sqrt{n^2 - 4}} =\)

\(\displaystyle \frac{4n}{n + \sqrt{n^2 - 4}} =\)

\(\displaystyle \frac{4}{1 + \sqrt{1 - \frac{4}{n^2}}}\)

now take the limit as \(\displaystyle n \to \infty\)