Question

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

Answer 1

See the explanation, below.

Explanation:

Preliminary Analysis

We want to make `abs((x^2-3x)-(-2)) < epsilon`

By choosing `delta`, we control the maximum size of `abs(x-2)`

We now look at `abs((x^2-3x)-(-2))` keeping in mind what we control.

`abs((x^2-3x)-(-2)) = abs(x^2-3x+2)`

`= abs((x-1)(x-2))`

`= abs(x-1)abs(x-2)`

If we knew the size of `abs(x-1)`, we could choose `abs(x-2)` to make sure that the product is `< epsilon`

Let's start by making sure that `x` is a little close to `2`, say `abs(x-2) < 1` (`delta` will need to be at most `1`.)

If `abs(x-2) < 1`, then `-1 < x-2 < 1`, that is: `1 < x < 3`.

If follows that `0 < x-1 < 2`, so we have `abs(x-1) < 2`

Now if we ALSO make sure that `abs(x-2) < epsilon/2`,

then we will have `abs(x-1)abs(x-2) < 2abs(x-2) < 2 (epsilon/2) = epsilon`

Now we are ready to write the proof:

Proof

Given `epsilon > 0`, let `delta = min{1,epsilon/2}`.
(Note that #delta is positive.)

For every `x` that satisfies `0 < abs(x-2) < delta`
we have

(`abs(x-1) < 2` and also)

`abs((x^2-3x)-(-2)) = abs(x^2-3x+2)`

`= abs((x-1)(x-2))`

`= abs(x-1)abs(x-2)`

`< (2)(epsilon/2) = epsilon`

We have shown that for any `epsilon > 0` there is a `delta > 0` such that

if `0 < abs(x-2) < delta`, then `abs((x^2-3x)-(-2)) < epsilon`.

By the definition of limit, `lim_(xrarr2)(x^2-3x) = -2`.