Prove with delta-epsilon definition: $\displaystyle\lim_{x\to 3} \frac{x^2-9}{x^2-3x}=2$
I started with the following inequality: $$ \Big|\frac{x^2-9}{x^2-3x}-2\Big|<\epsilon. $$ Then I could not progress after this: $$ \Big|\frac{x+3}{x}-2\Big|<\epsilon. $$ Could anyone help me to find $\delta=\delta(\epsilon)?$
$|\frac{x^2-9}{x^2-3x} -2| = |\frac{-x^2+6x-9}{x^2-3x}| = |\frac{x^2-6x+9}{x^2-3x}| = |\frac{x-3}{x}| = |x-3||\frac{1}{x}|$
You want $|x-3||\frac{1}{x}| < \epsilon$ when $|x-3|<\delta$ so we can bound $\delta \leq 1$ then $|x-3|<1$ so $|\frac{1}{x}| < \frac{1}{2}$ so we choose $\delta = \min\{1, 2\epsilon\}$ then $$|x-3||1/x| <\frac{1}{2}2\epsilon = \epsilon $$