How do you prove the statement lim as x approaches 1 for `(5x^2)=5` using the epsilon and delta definition?

Preliminary Analysis

We need to make `abs(5x^2-5) < epsilon` by making `abs(x-1) < delta`. (We choose the `delta`.)

So we begin by examining `abs(5x^2-5)`

`abs(5x^2-5) = abs(5(x^2-1)) = abs5abs(x^2-1)`

`= 5 abs((x+1)(x-1)) = 5 abs(x+1) abs(x-1)`

We control, through our choice of `delta`, the maximum for `abs(x-1)`.

If we make sure that `delta` is less than a number we choose now, then we can also control the size of `abs(x+1)`.
(If we put a bound on the distance between `x` and `1`, the we also get a bound on the distance between `x` and `-1`)

Any number will do, but is might be easiest for purpose of illustration to pick a number we haven't use yet so we can keep track of it.

Let's make sure that `delta <= 2`.

If `abs(x-1) < 2`, then `-2 < x-1 < 2`.

So, adding 2 to each part, we get, `0 < x+1 < 4` and `abs(x+1) < 4`.

We want `5 abs(x+1) abs(x-1) < epsilon` and we plan to make sure that `abs(x+1) < 4`, so we now need

`5 abs(x+1) abs(x-1)`

`"which is" < 5(4)abs(x-1)`
`"which is"= 20 abs(x-1)`
`"we want this" < epsilon`

So let's make sure that, in addition to `delta <= 2`,

we also want `delta <= epsilon/20`

Now we are ready to write the proof.

Proof
Claim: `lim_(xrarr1)5x^2 = 5`

Given `epsilon > 0`, let `delta = min{2, epsilon/20}` (Note that `delta > 0` as required.)

If `x` is chosen so that `0 < abs(x-1) < delta` then

note first that `abs(x+1) < 4`
( `abs(x-1) < delta <= 2 rArr -2 < x-1 < 2 rArr 0 < x+1 < 4`).

Furthermore, for such `x`, we have

`abs(5x^2-5) = 5abs(x^2-1) = 5abs(x+1)abs(x-1)`

`< 5(4) abs(x-1) = 20abs(x-1)`

`< 20(epsilon/20) = epsilon`.

That is: for `0 < abs(x-1< delta`, we have #abs(5x^2-5) < epsilon.

Therefore, by the definition on limit,

`lim_(xrarr1)5x^2 = 5`