Question

Using the $\in -\delta$ definition prove that $\underset{x\to 3}{lim}(x^2+2x-8)=7$

Answer 1

Suppose $f\left(x\right)=x^2+2x-8$

$\underset{x\to 3}{lim}f\left(x\right)=7$

For any value of $f>0$ s.t $0<|x-7|<0$

We need to find a value ef s.t

$0<|f\left(x\right)-7|<ef$

$|x^2+2x-8-7|<ef$

$\Rightarrow |x^2+2x-15|<ef$

$\Rightarrow |(x+5)(x-3)|<ef$

$\Rightarrow |x+5||x-3|<ef$

Since we are interest in the domain where $x\to 3$, we have a limit $x\in (1,4)$ and this area $|x+5|=x+5<7$

So, $|f\left(x\right)-7|=|x+5|\cdot |x-3|<7|x-3|$

We know that $|x-3|<f$

So $|f\left(x\right)-7|<f/7$

Pick $ef=f/7\Rightarrow |f\left(x\right)-7|<ef$ provided

$0<f<|x-3|$.