Limit of $(5n^2+2n)/(n^2-3)$ using limit definition

Question

I'm having trouble using the limit def to find this answer

I have to show $\lim (5n^2+2n)/(n^2-3) = 5$

so I have so far $|(5n^2+2n)/(N^2-3) - 5| < \epsilon$ so from there I got $|(2n+15)/(n^2-3)|<\epsilon$ so then $|(2+15/n)-(n-3/n) \leq 17/(n-3)$

From there I don't know what to do.

The answer I was given says that $n=(1/\epsilon + 3)$ then $n \ge N$ thus the inequality is true. But I'm not exactly sure why

Answer 1

First note $(5n^2+2n)/(n^2-3) > 5$ for all positive n.

So for an arbitrary $\epsilon$ we want it to be true that

$(5n^2+2n)/(n^2-3) - 5 < \epsilon$ for all large enough n. So we want to find just how large the n have to be for this to be true. We assume n > 0 so

$5n/(n^2 - 3) < (5n^2+2n)/(n^2-3) - 5 < \epsilon$

$5n^2 - 5(n^2-3) < (n^2-3) \epsilon$

$15 < (n^2 - 3)\epsilon$

$15/\epsilon +3 < n^2 $

So this will be true for all $n >= N >= \sqrt(15/\epsilon +3)$.