E-N Definition of Limit with a minus in the denominator I'm working on a problem, using E-N limits of which I have not seen an example of before, or could find an example of on the internet.
$$s_n = \frac{4n^2 + \sin(n)}{2n^2-1}$$
By inspection, $\lim\limits_{n\to\infty}s_n=2$.
$$s_n - 2 = \frac{4n^2 + \sin(n)}{2n^2-1} -2 = \frac{\sin(n) + 2}{2n^2 - 1}$$
The solution to this problem is not important, all I would like to know is how to deal with the $-1$ in the denominator, as every other problem has a $+1$ in and can be easily manipulated
Many thanks!
 A: Choose $N = \sqrt{\frac{3/\epsilon + 1}{2}}$.
Then $\forall n\in\mathbb{N}$ such that $n > N$,
$|\frac{\sin(n)+2}{2n^2-1}| \le |\frac{3}{2n^2-1}| < |\frac{3}{2(\sqrt{\frac{3/\epsilon + 1}{2}})^2-1}| = |\frac{3}{2(\frac{3/\epsilon + 1}{2})-1}| = |\frac{3}{3/\epsilon + 1-1}| = |\epsilon| = \epsilon$.
A: Oops, I just noticed you said the solution is not important. Then I will describe my process, which works for most of these types of problems.
Most humans (myself included) do scratch work before writing the proof. It goes like this.

*

*Go ahead and set $|s_n - L| < \epsilon$. This helps keep the goal in mind.

*Simplify the LHS (like you did).

*Use inequalities to get rid of as many things as you can. In this case I used $|\sin(x)| \le 1$. Even if the denominator was $|2n^2 + 1|$, I don't think we can eliminate it, because we'd end up with $3 < \epsilon$ (not helpful). We could have eliminated the $+1$, but that's a minor thing.

*Just solve for n. Remember to flip the inequality when you multiply by -1 or take the reciprocal of both sides.

*Turn that $n$ and $>$ into an $N$ and $=$. Time to start writing the proof. This is your N.

*Now for all $n\in\mathbb{N}$ such that $n>N$, consider $|s_n - L|$.

*Simplify it and all that stuff. $n$ should be in the denominator.

*Write a $<$ and plug that expression for $N$ into $n$.

*Everything should evaluate to $\epsilon$.

