<s>[INST] Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ I'm looking for assistance with the following problem:
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ where $f(x)=x^2-1$. 
My attempt:
We need $\delta$ such that $\vert f(x)-2 \vert < .2$. That is, we need $\vert x^2 - 1 - 3 \vert = \vert x^2 - 4 \vert = \vert x+2 \vert \cdot \vert x-2 \vert < .2$. We can control $\vert x - 2 \vert$ by $\delta$. Let us then restrict $x$ to the interval $(1,3)$. Thus $\vert x+2 \vert < 5$. Then we will let $\delta = \frac{.2}{5}$. 
Now then, if $0 < \vert x - 2 \vert < \delta = \frac{.2}{5}$ then $\vert f(x)-3 \vert =  \vert x^2 - 4 \vert = \vert x+2 \vert \cdot \vert x-2 \vert < 5 \cdot \frac{.2}{5}=.2.$ $\Box$
Does this work for a valid delta? I'm basically taking the more general case where $\delta$ would be $\min(\frac{\epsilon}{5},1)$ and trying to fit it to this specific case.
 [/INST] Since
\begin{eqnarray}
|f(x)-3|&=&|x^2-4|\\
&=&|x-2||(x-2)+4|\\
&\le& |x-2|(|x-2|+4)\\
&=&|x-2|^2+4|x-2| 
\end{eqnarray}
if we solve the inequality
$$
y^2+4y\le 0.2
$$
we get
$$
a\le y\le b,
$$
with
\begin{eqnarray}
a&=&-2-\sqrt{4.2}\approx-4.0494\\ 
b&=&-2+\sqrt{4.2}\approx0.0494\\ 
\end{eqnarray}
Hence, choosing $\delta \in (0,b)$, we have
$$
0<|x-2|<\delta \implies a<|x-2|<b \implies |f(x)-1|\le |x-2|^2+4|x-2|<0.2
$$
</s>