Expressing Mins/Maxs Using Limits: f(x) = 2x - 3 on interval [0,2)

[Eigendorf]

Hello again!

I'm trying to understand how people express min/max's when the min/max approaches an open interval.

This might be easier to explain with a function.

Lets say f(x) = 2x - 3 on the interval [0,2) closed interval at 0, open interval at 2

f'(x) = 1 so the min / max are just the end points of the interval.... but... the interval is open at 2.

The min for the interval is P(0,3) but I am confused on how to express the max for the interval.

Would I express it something like (X → 2 , Lim X → 2 f(x)) The max might not technically exist because you can take X as arbitrarily close to 2 as you wish, but I think it would still be useful to express the manner in which the max doesn't exist.

Am I over thinking this?

 

[mmm4444bot]

Your example function f has no local maximum, in the given interval.

As x approaches 2 from the left, the function values are:

0.9
0.99
0.999
0.9999
0.99999
0.999999 …

As this pattern continues forever, there is no maximum value.

We could take the limit of f(x), as x approaches 2, and we would find that f(x) approaches 1. However, this limiting value is not a function value because (in the limit) x never takes on the value 2.

The situation is similar to asking the questions, "What is the largest negative number?" or "What is the smallest positive number?"

There are no such numbers.