Question

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

Answer 1

See the explanation section below.

Explanation:

Our preliminary work

We need to show that, given any positive `edpsilon`, there is a `delta` (also positive) that guarantees that if `x` is chosen so that

`0 < abs(x-4) < delta`, then we will also have `abs((7-3x)-(-5))< epsilon`.

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

`abs((7-3x)-(-5)) = abs(7-3x+5)`

`= abs(12-3x)`

Because we say how to find `delta`, that means we control the sive of `abs(x-4)`

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

`abs((7-3x)-(-5)) = abs(12-3x) = abs(-3(x-4))`

`= abs(-3)abs(x-4) = 3abs(x-4)`

We have found that `abs((7-3x)-(-5))` which we want to make less that `epsilon`, is equal to `3abs(x-4)`, which is `3` times the thing we control.

If we make `abx(x-4) < epsilon/4`, that will work.

Now we need to present ourwokr as a proof.

Proof

Given `epsilon > 0`, choose `delta = epsilon/4`. Now if `x` is chosen so that `0 < abs(x-4) < delta`, then we have

`abs((7-3x)-(-5)) = abs(12-3x) = abs(-3)abs(x-4)`

`= 3abs(x-4) < 3 delta = 3(epsilon/3) = epsilon.`.

That is, `abs((7-3x)-(-5)) < epsilon.`.

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

`abs(x-4) < delta` implies that `3abs(x-4) < 3delta`.