<s>[INST] Proof that existsts $\delta>0$ such that $2-\frac{1}{3} < x^2 +x < 2+\frac{1}{3}$ I need to prove that exists:
$\delta>0$ such that $1-\delta < x < 1+ \delta\implies 2-\frac{1}{3} < x^2 +x < 2+\frac{1}{3}$
What I did:
$1-\delta < x < 1+ \delta\implies |x-1|<\delta$ and then I suppose that:
$$|x+2|<\delta$$
So:
$$|x-1||x+2|<\delta^2$$
And then I have:
$$|x^2 +x -2|<\delta^2$$
By setting: $\delta^2 = \frac{1}{3}$ we have $\delta=\frac{1}{\sqrt{3}}$ then:
$$|x^2 +x -2|<\frac{1}{3} \tag{for $\delta = \frac{1}{\sqrt{3}}$}$$
But then I graphed this and it didn't work. How, then, it could be done? Where's my error? Is it in supposing $|x+2|<\delta$ too?
 [/INST] We continue on the path you started. We want $|x-1||x+2|$ to be "small." The idea is to make $|x-1|$ small, that is, to choose $x$ close to $1$. 
Unfortunately, the $x+2$ term could spoil things. But for example if we make $|x-1|\lt \frac{1}{2}$, then $x+2$ will be between $2+\frac{1}{2}$ and $2+\frac{3}{2}$, and in particular will be positive and less than $4$. So if we make $|x-1|\lt \frac{1}{12}$, then $|x-1||x+2|$ will be less than $\frac{1}{3}$. 
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