Finding an epsilon-delta proof of a sequence

Question

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!

Answer 1

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 $\lim\limits_{x \to \infty} \frac{2x-3}{x^2-2x-5} = 0$