# A limit proof

1. Nov 3, 2012

### bedi

Prove that the limit of (3n+5)/2(n+1)^2 is 0 when n goes to infinity.

Attempt: I need to find an N such that for any €>0, (3n+5)/2(n+1)^2<€ holds for every n with n>N.

Then I made some manipulation;

(3n+5)/2(n+1)^2 < (3n+5)/(2n^2 +4n) < (3n+6)/(2n^2 +4n) = (n+3)/(n^2 +2n) < (n+3)/n^2

2. Nov 3, 2012

### tiny-tim

hi bedi!

(try using the X2 button just above the Reply box )

= n/n2 + 3/n2 ?

(btw, you could have started (3n+3)/2(n+1)2 + 2/2(n+1)2 )

3. Nov 3, 2012

### bedi

Alright, then should I ignore 3/n2?

4. Nov 3, 2012

### tiny-tim

nooo, you should prove that its limit is 0 !

5. Nov 3, 2012

### bedi

So I choose N such that 1/N<ε, which is permitted by the Archimedean property. Hence 1/n<1/N<ε. This proves that the limit of the first term is 0. To show that the limit of 3/n2 is also 0, I can use the same argument, can't I?

6. Nov 3, 2012

yup!