Simple epsilon-delta question Is it true that for every $ε > 0$, there is $δ > 0$, such that $0 < |x−2| < δ ⇒ |(x^2 −x)−2| < ε$?
Now I know that $|(x^2 −x)−2|$ is same as $|(x-2)(x+1)|$, but I am not sure how to link that with the first bit of info given. In general epsilon-delta proofs confuse me. 
So I start by saying that there is an epsilon s.t $|(x^2 −x)−2| < ε$. And if this is true then there is a delta s.t $0 < |x−2| < δ$. Or is it the other way around? 
Now, if $|(x^2 −x)−2| < ε$ then $|(x-2)(x+1)| < ε$ and $|x-2||x+1| < ε$ and
$$|x-2|<\frac{ε}{|x+1|}$$ But since epsilon is always positive and so is $|x+1|$ then a delta always exists. 
Is my proof correct or totally wrong? I feel as though all I have done is rearranged the equation, and not really proved anything. 
 A: Let the $\epsilon = \epsilon_0$ satisfying $|x^2-x-2| < \epsilon_0$. Initially choose $\delta$ to be $1$. We will refine this delta.
$-\epsilon_0 < x^2-x-2 < \epsilon_0$
$\implies -\epsilon_0+\frac{9}{4} < x^2-x+\frac{1}{4} < \epsilon_0+\frac{9}{4}$
$\implies -\epsilon_0+\frac{9}{4} < (x-\frac{1}{2})^2 < \epsilon_0+\frac{9}{4}$
$\implies -\epsilon_0+\frac{9}{4} < (x-\frac{1}{2})^2 $
Now, if $\epsilon_0 < \frac{9}{4}$
$\implies \sqrt{-\epsilon_0+\frac{9}{4}} < x-\frac{1}{2} $
$\implies \sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2} < x+1 $
Remember we had $|x^2-x-2| < \epsilon_0$. Then
$-\epsilon_0 < (x-2)(x+1) < \epsilon_0$
$\implies -\frac{\epsilon_0}{x+1} < (x-2) < \frac{\epsilon_0}{x+1}$
$\implies |x-2| < \frac{\epsilon_0}{x+1} < \frac{\epsilon_0}{\sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2}}$.
Therefore choose $\delta = min( 1 , \frac{\epsilon_0}{\sqrt{-\epsilon_0+\frac{9}{4}} + \frac{3}{2}} )$.
(note that for $\epsilon_0 > \frac{9}{4}$, choose $\delta$ as if $\epsilon_0 = \frac{9}{4}$, this will eventually satisfy the condition)
A: I always like to refer people to my answer here when it comes to simple polymonial $\delta$-$\epsilon$ proofs. Read this link so that you understand my methodology here.
Scratch work:
$$|x^2-x-2| = |(x-2)(x+1)| = |x-2||x+1|\text{.} $$
Take $\delta = 1$. Then
$$|x-2| < 1 \Longleftrightarrow -1 < x-2 < 1 \Longleftrightarrow 1 < x < 3 \Longleftrightarrow 2 < x+1 < 4 \implies |x+1| < 4\text{.}$$
So for $\epsilon > 0$,
$$|x-2||x+1| < 4|x-2| < 4\delta = 4\left(\dfrac{\epsilon}{4}\right)=\epsilon$$
if $\delta = \dfrac{\epsilon}{4}$, so we choose $\delta = \min\left(1, \dfrac{\epsilon}{4}\right)$.
Proof:
Let $\epsilon > 0$ be given. Choose $\delta = \min\left(1, \dfrac{\epsilon}{4}\right)$. Then
$$|(x^2-x)-2| = |x^2-x-2| = |x-2||x+1| < 4|x-2|$$
(since if $|x-2| < 1$, $|x+1| < 4$), and 
$$4|x-2| < 4\delta \leq 4\left(\dfrac{\epsilon}{4}\right) = \epsilon\text{.}$$
A: Here are some general results which enable us to handle the Q of continuity for a broad class of real functions: 
(1)... Constant functions are continuous.
(2)... f(x)=x is continuous 
(3)...f(x)=|x| is continuous.
(4)...For continuous f, g :
...  (i)... h(x)=f(x)+g(x) is continuous.
...  (ii)...j(x)=f(x)g(x) is continuous.
...  (iii)...k(x)=f(g(x)) is continuous. 
These are readily proven by the standard $\epsilon , \delta$ method. One immediate consequence is that a polynomial $p(x)$ and its absolute value $|p(x)|$ are continuous functions
