<s>[INST] How do I solve delta epsilon proofs for quadratic equations? For the $\lim\limits_{x\rightarrow 2} (x^2 + 5x - 2) = 12$ I need to show how to find a $\delta$ such that $|f(x) - L| < \varepsilon$ for all $x$ satisfying $0 < |x - a| < \delta$.
Help is appreciated.
 [/INST] We have show that for all $\epsilon>0$ there is a $\delta >0$ such that 
$$
|x-x_0|<\epsilon \implies | (x^2+5x-2) - 12|<\epsilon .
$$
We will manipulate the expression $| (x^2+5x-2) - 12|<\epsilon$ to find a candidate to $\delta$. 
\begin{align}
|  (x^2+5x-2) - 12|<\epsilon \implies & -\epsilon <x^2+5x-14<\epsilon \\
\implies& -\epsilon <\left(x-\frac{5}{2}\right)^2 -\frac{25}{4}-14<\epsilon \\
\implies& \frac{81}{4}-\epsilon <\left(x-\frac{5}{2}\right)^2<\frac{81}{4}+\epsilon \\
\implies& \sqrt{\frac{81}{4}-\epsilon} <\left|x-\frac{5}{2}\right|<\sqrt{\frac{81}{4}+\epsilon} \\
\implies& \left|x-\frac{5}{2}\right|<\sqrt{\frac{81}{4}+\epsilon} \\
\implies& - \sqrt{\frac{81}{4}+\epsilon}<x-\frac{5}{2}<\sqrt{\frac{81}{4}+\epsilon} \\
\implies& \frac{1}{2}- \sqrt{\frac{81}{4}+\epsilon}<x-2<\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2} \\
\implies& |x-2|<\max\left\{\left|-\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2}\right|,\left| +\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2}\right| \right\} \\
\end{align}
Set  $\delta=\max\left\{\left|-\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2}\right|,\left| +\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2}\right| \right\}$. In the above inequalities there is no loss of generality if we fix $\epsilon < \frac{81}{4}$. Then
\begin{align}
0<|x-2|<\delta \implies & -\delta < x-2 <+\delta \\
\implies& \frac{1}{2}- \sqrt{\frac{81}{4}+\epsilon}<x-2<\sqrt{\frac{81}{4}+\epsilon}+\frac{1}{2} \\
 \vdots\quad  & \\
   & \\
\vdots \quad & \\
\implies & | (x^2+5x-2) - 12|<\epsilon
\end{align}
</s>