Question

How do you prove limit of `x^2-3x=-2` as `x->2` using the precise definition of a limit?

Answer 1

Please see the explanation section below.

Explanation:

Given `epsilon > 0` choose `delta = min{1, epsilon/3}`. Note that `delta > 0`

For every `x` with `0 < abs(x-2) < delta`, we have

`abs(x-2) < 1`, so `-1 < x-2 < 1` and `1 < x < 3`.

Now `abs((x^2-3x)-(-2)) = abs(x^2-3x+2) = abs(x-2)abs(x-1)`

And if `abs(x-2) < delta` then `abs(x-2) < epsilon/3` and `abs(x-1) < 3`, so

So, if `abs(x+2) < delta`, then

`abs((x^2-3x)-(-2)) = abs(x+2)abs(x^2-2x+4)`

`< (epsilon/3)*(3) = epsilon`.

We have shown that for any positive `epsilon` there is a positive `delta` such that for any `x` with `0 < abs(x-2) < delta`, we have `abs((x^2-3x)-(-2)) < epsilon`.

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