Stuck with proving $\lim_{x\to 2} (x^2-3x)=-2$, using the $\epsilon, \delta$ deifnition I was asked to prove that: $\lim_{x\to 2} (x^2-3x)=-2$ using the $\epsilon, \delta$ deifnition 
I started to try and solve the epsilon inequality in this manner: 
for every  $\epsilon>0$ there exist $\delta >0$ such that if:    
$0<|x-2|<\delta$ then $|(x^2-3x)-(-2)|<\epsilon$
Then I started to manipulate my epsilon inequality in order to get a delta in terms of epsilon by doing the following:
$$|(x^2-3x)-(-2)|=|x^2-3x+2|=|(x-2)(x-1)|=|x-2||x-1|$$
Now, we must see that 
$$|x-1|=|x-2+1| ≤ |x-2|+1$$
And thus, that:
$$|x-2||x-2+1|≤ (|x-2|+1)|x-2|$$
After this step I get stuck. I do not know how to proceed or how to use the equations I have in order to get delta in terms of epsilon.
Thanks in advance for the help. 
 A: You're almost there.  If $\delta \leq 1$, and $|x-2|<\delta$, then you know 
$$
    |x-1| \leq |x-2| +1 \leq 2
$$
and also
$$
    |x-1||x-2| < 2 \delta
$$
In order for the right-hand side to be $\leq\epsilon$, you need to make sure that $\delta \leq \frac{\epsilon}{2}$.  In order to guarantee both $\delta \leq 1$ and $\delta \leq \frac{\epsilon}{2}$, choose $\delta = \min\left\{1,\frac{\epsilon}{2}\right\}$.
A: Since your $x$ is approaching $2$, you may assume $|x-2|< 1/2$ by choosing a positive  $\delta <1/2$
in that case you have $3/2<x<5/2$ and as a result $1/2<x-1<3/2$ 
so $|x-1|<3/2$
Now if $|x-2|<\delta$  then $|(x-2)(x-1)| < (3/2)\delta$
Thus if your $\delta = \min \{1/2, 2\epsilon/3\}$, then $$ |x-2|<\delta \implies |   (x-2)(x-1)| <\epsilon$$
A: You can continue as follows:
$$|x-2||x-2+1|≤ (|x-2|+1)|x-2|<(\delta+1)\delta=\epsilon \Rightarrow \\
\delta^2+\delta-\epsilon=0 \Rightarrow \delta =\frac{-1+\sqrt{1+4\epsilon}}{2}.$$
Hence, for the given $\epsilon>0$, you can choose $\delta=\frac{-1+\sqrt{1+4\epsilon}}{2}>0$, so that when $0<|x-2|<\delta$, $|(x^2-3x)-(-2)|<\epsilon$. 
