Question

prove that lim as x approaches 3 of (x^2+x-4)=8 using epsilon and delta. explain steps please

Answer 1

\[\begin{align*}\left|\left(x^2+x-4\right)-8\right|&=\left|x^2+x-12\right|\\
&=\left|(x-3)(x+4)\right|\\
&=|x-3||x+4|
\end{align*}\]
Suppose we agree to set \(\delta\le1\), then
\[\begin{align*}
|x-3|<1~~\implies&~~-1<x-3<1\\
&~~~~~~~6<x+4<8\\
&~~~~~~~~~~~~|x+4|<8
\end{align*}\]
Then we have
\[\begin{align*}\left|\left(x^2+x-4\right)-8\right|&=|x-3||x+4|\\
&<8|x-3|
\end{align*}\]
So for this to be less than \(\epsilon\), we can set \(\delta=\min\left\{1,\dfrac{\epsilon}{8}\right\}\)