Question

How do you prove the statement lim as x approaches 2 for `((x^2+x-6)/(x-2))=5` using the epsilon and delta definition?

Answer 1

See the explanation below.

Explanation:

Preliminary Analysis

First look at `abs(f(x)-L)`. I want to make that less than `epsilon`.

I am looking for `abs(x-2)` because, through my control of `delta`, I control `abs(x-2)`.

`abs(f(x)-L) = abs((x^2+x-6)/(x-2)-5)`

`= abs(((x+3)(x-2))/(x-2)-5)`

`= abs((x+3)-5)` for all `x != 2`

`= abs(x-2)` for all `x != 2`

We want to make `abs(f(x)-L)` less than a given `epsilon`.

If we make `abs(x-2) < epsilon` BUT `x != 2`, then we will have what we want. (We cannot allow `x=2` because `2` is not in the domain of `f`.

So now, we are ready to write the proof.

Proof

Given `epsilon > 0` choose `delta = epsilon`. (Clearly, then, `delta > 0` as required.

Now if `x` is chosen so that

`0 < abs(x-2) < delta`, then we will have

`abs((x^2+x-6)/(x-2)-5) = abs(((x+3)(x-2))/(x-2)-5)`

`= abs((x+3)-5) = abs(x-2) < delta = epsilon`.

That is:
If `0 < abs(x-2) < delta`, then `abs((x^2+x-6)/(x-2)-5) < epsilon`.

By the definition of limit,

`lim_(xrarr2)(x^2+x-6)/(x-2) = 5`