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.

The other solution is of course neat :-)
If $|x^2-4|<0.2$ then we have that $3.8<x^2<4.2$ that is either 


*

*$\sqrt{3.8}<x<\sqrt{4.2}$ or

*$-\sqrt{4.2}<x<-\sqrt{3.8}$.
Consider case $1$. From here we get 
\begin{align}
-0.05064\sim\sqrt{3.8}-2&<x-2<\sqrt{4.2}-2\sim0.04939
\end{align}
Now if we want $-\delta<x-2<\delta$, then we can choose $\delta<\sqrt{4.2}-2$.
For case $2$ we have 
\begin{align}
-4.04939\sim-\sqrt{4.2}-2&<x-2<-\sqrt{3.8}-2\sim -3.94936
\end{align}
implying that $\delta$ shall be negative which is not possible.
