$\lim_{x\to0}\frac{x^2+2x}{x^2-3x} = - \frac{2}{3}$
Given that $\epsilon = 0.01$, what is the largest possible $\delta$ according to the epsilon-delta definition of the limit?
Here's what I've done by myself, but I'm getting stuck:
$\left|\frac{x^2+2x}{x^2-3x}+\frac{2}{3}\right|<\epsilon\ \text{ if }0
I could split the absolute value into two separate ones and multiply the denominator up to the other side, but that wouldn't get me any closer to isolating x. Even plugging in 0.01 for $\epsilon$ doesn't seem to help. What am I doing wrong?