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

2 Answers
Apr 6, 2016

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#.

Apr 6, 2016

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.