Limit proof using epsilon-delta So I need to prove $$\lim_{x \to 2} \frac{2x^2-3x-2}{x^2-5x+6}.$$
Here's what I did:
$$\left|\frac{2x+1}{x-3} + 5\right| = \left|\frac{7(x-2)}{x-3}\right| < 7\delta\left|\frac{1}{x-3}\right|,$$
and now I am stuck; normally we will further restrict delta, then choose the minimum of the two. However, delta is always negative no matter how I restrict it.
 A: In this kind of exercise, I often see people get confused about the limit not being in zero.
My constant advice is to write $x=2+u$ with $u\to 0$.
For the $\varepsilon,\delta$ proof, the condition is then $|x-2|=|u|<\varepsilon$
So let's begin by calculating 
$f(2+u)=\dfrac{2(2+u)^2-3(2+u)-2}{(2+u)^2-5(2+u)+6}=\dfrac{8+8u+2u^2-6-3u-2}{4+4u+u^2-10-5u+6}=\dfrac{2u^2+5u}{u^2-u}=\dfrac{2u+5}{u-1}$
And this has a limit $\ell=-5$ when $u\to 0$.
Next step is to write $|f(x)-\ell|=\left|\dfrac{2u+5}{u-1}+5\right|=\left|\dfrac{7u}{1-u}\right|$
This is where this expression is easier to manipulate than $\dfrac{7(x-2)}{x-3}$


*

*we have to find an upper bound for the numerator: $|7u|<7\varepsilon$

*and a lower bound for the denominator: $|1-u|>\frac 12$ for $|u|<\frac 12$


Now select $\delta=\min(\varepsilon,\frac 12)$ such that both inequalities above are verified when $|u|<\delta$.
Finally you get: $$|x-2|<\delta\implies |f(x)-\ell|<14\varepsilon$$

Have a look at another $\varepsilon,\delta$ proof there $\to$ Delta Epsilon proof of limit (how to find which numbers to plug in and how to derive delta min?)
A: If you choose $\delta <1/2$ then $\left|x-3\right|>1/2$ so 
$$\left|\frac{7\left(x-2\right)}{x-3}\right|<14\delta$$
A: Consider $|x-2|<0.5$ , then,
$-1.5 <x-3 < -0.5$, or 
$1.5 >-(x-3) >0.5$, or
$|x-3| >1/2.$
Let $\epsilon > 0$ be given.
Choose $\delta \le \min (1/2, \epsilon/(14))$.
Then $|x-2| \lt \delta$ implies
$|\dfrac{2x+1}{x-3} -(-5)| =|\dfrac{7(x-2)}{x-3}| \lt$
$7\delta |\dfrac{1}{1/2}| =14\delta \lt \epsilon$.
