Proving the Limit of (3n+5)/2(n+1)^2 is 0 as n Approaches Infinity

[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

Then what? Please help.

 

[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 )

 

[bedi]

Alright, then should I ignore 3/n²?

 

[tiny-tim]

Alright, then should I ignore 3/n²?

nooo, you should prove that its limit is 0 !

 

[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?

 

[tiny-tim]

yup!