# How do you prove the statement lim as x approaches 3 for (x^2+x-4) = 8 using the epsilon and delta definition?

Sep 19, 2015

#### Explanation:

We want to make $\left\mid \left({x}^{2} + x - 4\right) - 8 \right\mid < \epsilon$

We control $\delta$ and $\left\mid x - 3 \right\mid < \delta$

Note that:

$\left\mid \left({x}^{2} + x - 4\right) - 8 \right\mid = \left\mid {x}^{2} + x - 12 \right\mid$

$= \left\mid \left(x + 4\right) \left(x - 3\right) \right\mid$

$= \left\mid x + 4 \right\mid \left\mid x - 3 \right\mid$

We control $\left\mid x - 3 \right\mid$ and so, indirectly, we control $\left\mid x + 4 \right\mid$

Make $\left\mid x - 3 \right\mid < 1$ . (Any positive number will work in place of $1$, but the details of what follows will change.)

This assures us that $2 < x < 4$.

So considering $x + 4$ ( the other factor), we see:

$6 < x + 4 < 7.$ And

$\left\mid x + 4 \right\mid < 7$.

So, if $\left\mid x - 3 \right\mid < 1$, then

$\left\mid \left({x}^{2} + x - 4\right) - 8 \right\mid = \left\mid x + 4 \right\mid \left\mid x - 3 \right\mid < 7 \left\mid x - 3 \right\mid$

If we also make sure that $\left\mid x - 3 \right\mid < \frac{\epsilon}{7}$, then we will have:

$\left\mid \left({x}^{2} + x - 4\right) - 8 \right\mid < 7 \left\mid x - 3 \right\mid < 7 \left(\frac{\epsilon}{7}\right) = \epsilon$.

Now write the proof :

Given $\epsilon > 0$, let $\delta = \min \left\{1 , \frac{\epsilon}{7}\right\}$.

Now is $\left\mid x - 3 \right\mid < \delta$, we will have:

$\left\mid \left({x}^{2} + x - 4\right) - 8 \right\mid = \left\mid {x}^{2} + x - 12 \right\mid = \left\mid x + 4 \right\mid \left\mid x - 3 \right\mid$

$< 7 \left(\frac{\epsilon}{7}\right) = \epsilon$