<s>[INST] Finding an epsilon-delta proof of a sequence I'd like to write an epsilon-delta proof that
$$
\lim_{x \to \infty} \frac{2x-3}{x^2-2x-5} = 0 
$$
so what I've done is,
$\lvert \frac{2x-3}{x^2-2x-5} \rvert < \frac{2x}{x^2-2x-5}$ if $x>3/2$.
$$
\frac{2x}{x^2-2x-5} < \frac{2x}{x^2-7x} < \epsilon
$$
$$ 
x > \frac{2}{\epsilon} +7
$$
$
\therefore \lvert \frac{2x-3}{x^2-2x-5} \rvert < \epsilon $ if $ x>N(\epsilon) = Max [{\frac{3}{2}, \frac{2}{\epsilon}+7}]
$
Is this valid? Or is there a mistake that I made? If so, could I get an explanation? Thank you in advance!
 [/INST] You are essentially correct, though you are going to want $x > \sqrt{6}+1$ to avoid division by $0$ and to ensure $\frac{2x-3}{x^2-2x-5}>0$ Later you will want $x >7$ for the same reason. and since $\frac2\epsilon+7>7>\sqrt{6}+1> \frac32 >1$ you can slightly simplify your $N(\epsilon)$ expression. While your exploration is is sensible, a proof might better be in the opposite order.
So perhaps something like:

For any given $\epsilon>0$ and $x > N(\epsilon) = \frac2\epsilon+7$
you have $x>7$ and $x^2>7x$ and $5x > 5$ and $2x > 3$ and $x^2-2x-5>0$
in which case  $0 <  \frac{2x-3}{x^2-2x-5} < \frac{2x}{x^2-2x-5} < \frac{2x}{x^2-7x} = \frac{2}{x-7}\lt \epsilon$
so $\lim\limits_{x \to \infty} \frac{2x-3}{x^2-2x-5} = 0$

</s>