<s>[INST] 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. 
 [/INST] 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)
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