# How do you prove the statement lim as x approaches 4 for (7 – 3x) = -5 using the epsilon and delta definition?

Oct 7, 2015

See the explanation section below.

#### Explanation:

Our preliminary work

We need to show that, given any positive $e \mathrm{dp} s i l o n$, there is a $\delta$ (also positive) that guarantees that if $x$ is chosen so that

$0 < \left\mid x - 4 \right\mid < \delta$, then we will also have $\left\mid \left(7 - 3 x\right) - \left(- 5\right) \right\mid < \epsilon$.

Let's look at the thing we need to make less than $\epsilon$.

$\left\mid \left(7 - 3 x\right) - \left(- 5\right) \right\mid = \left\mid 7 - 3 x + 5 \right\mid$

$= \left\mid 12 - 3 x \right\mid$

Because we say how to find $\delta$, that means we control the sive of $\left\mid x - 4 \right\mid$

Not that if we factor out a $-$, we can get

$\left\mid \left(7 - 3 x\right) - \left(- 5\right) \right\mid = \left\mid 12 - 3 x \right\mid = \left\mid - 3 \left(x - 4\right) \right\mid$

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

We have found that $\left\mid \left(7 - 3 x\right) - \left(- 5\right) \right\mid$ which we want to make less that $\epsilon$, is equal to $3 \left\mid x - 4 \right\mid$, which is $3$ times the thing we control.

If we make $a b x \left(x - 4\right) < \frac{\epsilon}{4}$, that will work.

Now we need to present ourwokr as a proof.

Proof

Given $\epsilon > 0$, choose $\delta = \frac{\epsilon}{4}$. Now if $x$ is chosen so that $0 < \left\mid x - 4 \right\mid < \delta$, then we have

$\left\mid \left(7 - 3 x\right) - \left(- 5\right) \right\mid = \left\mid 12 - 3 x \right\mid = \left\mid - 3 \right\mid \left\mid x - 4 \right\mid$

$= 3 \left\mid x - 4 \right\mid < 3 \delta = 3 \left(\frac{\epsilon}{3}\right) = \epsilon .$.

That is, $\left\mid \left(7 - 3 x\right) - \left(- 5\right) \right\mid < \epsilon .$.

Note that
near the middle we used, without mention, the fact that

$\left\mid x - 4 \right\mid < \delta$ implies that $3 \left\mid x - 4 \right\mid < 3 \delta$.