Studying for exam, need help!!

[johnnyICON]

Hi can someone help me out. My girlfriend has an exam tomorrow and she got stuck on this question. Her professor decided not to give out any solutions she's not too sure if she is heading in the right direction. Any help would be great, thank you in advance.

Here is the question:
Prove by using the definition of the limit of a sequence that:
\lim_{n \to \infty} \frac{n + 1}{n^2} + 3 = 3

[mattmns]

split up the fraction


\frac{n}{n^2} + \frac{1}{n^2}

[johnnyICON]

How do I get just one term of n?

[matt grime]

You must show that given e>0 there is an N such that n>N implies

\frac{n+1}{n^2}<e

agreed?

Well,


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


so pick N such that N+1>1/e

[johnnyICON]

Second question, if you could get back to me asap, we're at school cramming right now
Leta_n = \frac{n^2-1}{2n^2+3} Prove by using the definition of the limit of a sequence that \lim_{n \to \infty}a_n = \frac{1}{2}

[matt grime]

Well, have you simplified a_n -1/2?

every question like this reduces to showing something tends to zero. that thing tends to zero for obvious reasons just like the previous example

[johnnyICON]

I just have |\frac{n^2-1}{2n^2+3} - \frac{1}{2}| and I don't know how to get it to a single n term.

[johnnyICON]

Thanks for your help anyway. We gotta go now.