Question

Prove that if f is continuous on [a, inf) and \[\lim_{x \rightarrow \inf}f(x) = A\]then f is bounded on [a, inf)

Answer 1

So I know that for the definition of a limit as x tends to infinity we have the following:\[(\forall \epsilon > 0)(\exists N):(x>N), |f(x)-A| < \]

Answer 2

\[\epsilon \]

Answer 3

Here goes - proof by contradiction.
Suppose the conclusion is false ===> there is a sequence of x-s that on this sequence the function tends to infinity (or to - infinity)

Answer 4

Either the sequence itself (which contains infinite number of points) is bounded and then it must have a final partial limit X_1. Or it is unbounded ==> then for arbitrarily large x we get larger and larger f(x_i) values.

Answer 5

Since the values of f are unbounded by the negation assumption, we get a contradiction - in case 1 with continuity, and in case 2 (unbounded sequence) with existence of a limit at infty.
Q.E.D.

Answer 6

When you wrote "x-s", was it meant to me x_s indicating a subsequence?

Answer 7

Well you might say . Better is the exact\[\left\{ x_i \right\}_{i=1}^{\infty}\]

Answer 8

So that sequence is in [a, inf), right?

Answer 9

This goes without saying - of course. But concentrate on the 2 cases of the sequence behaviour

Answer 10

So you said that there is a sequence x_n which is in [a, inf) such that the values of x_n tend to inf as n tends to inf, correct?

Answer 11

And then you said "either the sequence itself is bounded..." which negates what was just shown, doesn't it?

Answer 12

No - what I said was " such as that the values of f(x_n) ----> +infty or - infty

Answer 13

Sorry explaining every notion and word is too consuming.

Answer 14

No worries, just trying to get my head around it :)

Answer 15

So if the f(x_n) tends to +infty or - infty, then x_n is bounded?

Answer 16

Look I have provided COMPLETE proof by negation. The intuition is as follows:
If we suppose the function is unbounded then it must "explode" either in finite region or at infty. If it does so in finite region - there will be a pint of discontinuity.(contradiction to data) and if it does so on the way to x--> infty then this is a contradiction to existence of a finite limit of f. That' s all the proof.

I deserve a medal, believe me. Whether you realize it is the same as understanding the complete argument I've provided here.

Answer 17

Good bye and best of luck.