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.
 A: $\forall\epsilon >0\ \exists\delta>0\ $ such that $|x-a|<\delta\implies |f(x)-L|<\epsilon$.  Then you have, $|x-2|<\delta\implies |x^2+5x-2-12|<\epsilon$.  Factoring the right side of $\implies$
$$|(x-2)(x+7)|<\epsilon\implies |x-2|<\frac{\epsilon}{|x+7|}$$. Now we need to get the fraction on the right hand side only in terms of $\epsilon$ and some number.  You can set $|x-2|<1,$ because the limit only applies as $x\to 2,$ so you can restrict $\epsilon$ to being $<1$ so $|x-2|<1$.  Then this gives $-1<x-2<1\implies 1<x<3\implies 8<x+7<10.$  Since the maximum of $|x+7|$ makes $\frac{\epsilon}{|x+7|}$ a miniumum, then $$|x-2|<\frac{\epsilon}{|x+7|}<\frac{\epsilon}{10}.$$  Now you can let $\delta=\min(1,\frac{\epsilon}{10}).$  Then, just to show you if you let $\delta=1$, then $|x-2|<\delta =1\implies |x-2||x+7|<|x+7|.$  Since you know that $1<\frac{\epsilon}{10}$, then this means that $\epsilon<10.$  Therefore, $|x-2|x+7|<|x+7|<10<\epsilon\implies |(x-2)(x+7)|<\epsilon\implies|x^2-5x-14|<\epsilon.$ The proof for $\delta=1$ is done.  Now you can prove it for $\delta=\frac{\epsilon}{10},$ and once you do that the proof is done in entirety.  I'll leave that last part up to you for when $\delta=\frac{\epsilon}{10}$.
A: 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}
A: Let e > 0 be given, choose d = min{1, e/10}. So if 0 < /x - 2/ < d ===> /x - 2/ < 1 ===> -1 < x - 2 < 1 ===> 8 < x + 7 < 10 ===> /x + 7/ < 10 ===> / (x^2 + 5x  - 2) - 12 / = / x^2 + 5x - 14 / = /(x - 2)(x + 7)/ = /x + 7//x - 2/ < 10(e/10) = e. This means f(x) --> 12 as x --> 2. Here f(x) = x^2 + 5x - 12.
