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 $|f(x) - L| &lt; \varepsilon$ for all $x$ satisfying $0 &lt; |x - a| &lt; \delta$.

Help is appreciated.
$\forall\epsilon &gt;0\ \exists\delta&gt;0\ $ such that $|x-a|&lt;\delta\implies |f(x)-L|&lt;\epsilon$.  Then you have, $|x-2|&lt;\delta\implies |x^2+5x-2-12|&lt;\epsilon$.  Factoring the right side of $\implies$
$$|(x-2)(x+7)|&lt;\epsilon\implies |x-2|&lt;\frac{\epsilon}{|x+7|}$$. Now we need to get the fraction on the right hand side only in terms of $\epsilon$ and some number.  You can set $|x-2|&lt;1,$ because the limit only applies as $x\to 2,$ so you can restrict $\epsilon$ to being $&lt;1$ so $|x-2|&lt;1$.  Then this gives $-1&lt;x-2&lt;1\implies 1&lt;x&lt;3\implies 8&lt;x+7&lt;10.$  Since the maximum of $|x+7|$ makes $\frac{\epsilon}{|x+7|}$ a miniumum, then $$|x-2|&lt;\frac{\epsilon}{|x+7|}&lt;\frac{\epsilon}{10}.$$  Now you can let $\delta=\min(1,\frac{\epsilon}{10}).$  Then, just to show you if you let $\delta=1$, then $|x-2|&lt;\delta =1\implies |x-2||x+7|&lt;|x+7|.$  Since you know that $1&lt;\frac{\epsilon}{10}$, then this means that $\epsilon&lt;10.$  Therefore, $|x-2|x+7|&lt;|x+7|&lt;10&lt;\epsilon\implies |(x-2)(x+7)|&lt;\epsilon\implies|x^2-5x-14|&lt;\epsilon.$ 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 $\delta=\frac{\epsilon}{10}$.
Let e > 0 be given, choose d = min{1, e/10}. So if 0 &lt; /x - 2/ &lt; d ===> /x - 2/ &lt; 1 ===> -1 &lt; x - 2 &lt; 1 ===> 8 &lt; x + 7 &lt; 10 ===> /x + 7/ &lt; 10 ===> / (x^2 + 5x  - 2) - 12 / = / x^2 + 5x - 14 / = /(x - 2)(x + 7)/ = /x + 7//x - 2/ &lt; 10(e/10) = e. This means f(x) --> 12 as x --> 2. Here f(x) = x^2 + 5x - 12.