Question

How do you use the formal definition of a limit to prove `lim (x^2-x+1)=1` as x approaches 1?

Answer 1

Any polynomial f(x) with real (single valued) coefficients is single valued. The quadratic has the value 1 at x= 1. Only for discontinuous fuctions, we have limiting values for different approachesi

Explanation:

The question of passing on to the limit, if, any, does not arise for continuous single-valued functions. The quadratic is a continuous and single-valued function..

In general, the formal way for finding the limit is to substitute x = 1 + h and take the limit as `h to 0`.

Answer 2

See below.

Explanation:

Preliminary work

We need to show that
for any positive `epsilon`, there is a positive `delta` that makes the following true:

For every `x` with `0 < abs(x-1) < delta`, we get `abs((x^2-x-1)-1) < epsilon`.

Note that `abs((x^2-x-1)-1) = abs(x^2-x) = abs(x(x-1)) = absxabs(x-1)`

We start by putting an initial bound on `abs(x-1)`, say `abs(x-1) < 2`.

With this initial requirement, we see that `x` is within distance `2` of `1` (on the number line). Therefore, with this initial requirement, we get `x` is between `-1` and `3`. That is `-1 < x < 3`. And this entails that `absx < 3`.

Now if we ALSO make sure that `abs(x-1)` is less than `epsilon/3`, then we will be able to show that `absxabs(x-1) < (3)(epsilon/3) = epsilon` as we need.

Here then is the proof :

Given `epsilon > 0`, choose `delta = min{2, epsilon/3}`

Now if `0 < abs(x-1) < delta` then

a) `absx < 3`, and

b) `abs(x-1) < epsilon/3`

So

`abs((x^2-x-1)-1) = absxabs(x-1) < (3)(epsilon/3) = epsilon`

This completes the proof.