# Confusing limit

1. Dec 8, 2004

### danne89

My book tolds me that: $$\lim_{n\rightarrow\infty} \frac{n(n+1)}{2n^2}= \frac{1}{2}$$. I don't get it. Maybe this with infinity, I dunno... Please, help!

2. Dec 8, 2004

### quasar987

Because

$$\frac{n(n+1)}{2n^2} = \frac{n^2+n}{2n^2} = \frac{n^2}{2n^2} +\frac{n}{2n^2} = \frac{1}{2} + \frac{1}{2n}$$

3. Dec 8, 2004

### danne89

Ok, I think I got it now. So it's based on that 1/n equals 0, when n approaches infinity and 2 * infinity = infinity, right?

4. Dec 8, 2004

### fourier jr

yeah, the 1/(2n)=(1/2)(1/n) so you can factor the constant out of the limit, ie lim (1/(2n)) = (1/2) * lim (1/n) --> 0 as n --> infinity