Real Analysis proof limits and bounded functions

[kbrono]

Homework Statement

Let f be a function and p\in . Assume that a\leqf(x)\leqb near p. Prove that if L= lim f(x) as x-->p Then L\in [a,b]

The Attempt at a Solution

I want to say that because f(x) is bounded by [a,b] that automatically implies that the Limit L is also bounded by [a,b] and is therefore an element. But i have a feeling I'm supposed to make a sequence from the Sequential Characterization of Limits...

 

[Dick]

I think the easiest approach is to do a proof by contradiction. Assume L is not in [a,b] and derive a contradiction. Funny, I think I gave the same advice yesterday.

 

[kbrono]

Ah ok, thank you

 

[kbrono]

Ok here's what I tried

Basically I said assume L is not an element of [a,b] Then since f(x) is only defined on the interval [a,b]/{p} Then [a,b]/{p} contains L. Therefore L is an element of [a,b]

 

[╔(σ_σ)╝]

A quick an "cheap" way to do this is by subtracting L from all sides of the inequality and then take the limit of each side. Finally, rearrange the resulting inequality.