Question

Prove using epsilon,delta definition of limit.
lim (5x + 8) = 3 as x -> -1

Answer 1

We want to satisfy \(|(5x+8)-3|<\epsilon\) for all \(\epsilon>0\), by giving a value of \(\delta\) that will guarantee this when \(0<|x-(-1)|=\color{red}{|x+1|}<\delta\).

In order to find this \(\delta\), we do some algebra:
\[\begin{align*}
|(5x+8)-3|&=|5x+5|\\
&=|5||x+1|\\
&=5|x+1|
\end{align*}\]
Since we want to guarantee that this quantity is less than \(\epsilon\), we must have
\[\color{red}{|x+1|}<\frac{\epsilon}{5}\]
For this reason, we would let \(\delta=\dfrac{\epsilon}{5}\). The proof follows from there.