Proving the One Sided Limit Theorem: A Rigorous Approach

[RPierre]

Homework Statement

Prove that the limit as x->c of f(x) = L if and only if both one sided limits also = L

Homework Equations

Has to be an epsilon delta proof

The Attempt at a Solution

Being an if and only if, I have to do two cases : If A, then B. and if NOT A, then NOT B, logically.

Case 1:
Let lim x->c from the left be L, and lim x->c from the right be L.

then if $$c - \delta < x < c then |f(x) - L| < \epsilon$$

and if $$c < x < c + \delta then |f(x) - L| < \epsilon$$

Case 2:

Let lim x->c from the left = M, and lim x->c from the right = N.

This is all I have really rationalized I am kind of stumped how to do a rigorous proof of this, I.e. I know how to do specific proofs but not a rigorous general proof. \

Can anyone offer any help / a starting point =/ ?

 

[JG89]

Okay, we have two things to prove: 1) if f(x) goes to L as x goes to c, then both one sided limits also approach L. 2) If both one sided limits go to L for x approaching c, then f(x) goes to L as x goes to c.

For 1) we know that for every positive epsilon, there exists a positive delta such that $$|f(x) - L| < \epsilon$$ if $$|x - c| < \delta$$. We can rewrite $$|x-c| < \delta$$ as $$c - \delta < x < c + \delta$$. Now, say we're examining the right handed limit of f(x) as x->c, then we only look at the interval $$x < c + \delta$$, right? And with the left handed limits we only look at the interval $$c - \delta < x$$. You should be able to prove it now.

2) Look at the delta-interval of both one-sided limits and "put them together".

 

[scouter]

Is it possible to continue the solution? I can find the (1).

 

[JG89]

For 2), you're going to assume that both one-sided limits exist. So for every positive epsilon, there is a positive delta such that $$c - x < \delta \Rightarrow |f(x) - L| < \delta$$ and $$x - c < \delta \Rightarrow |f(x) - L|$$.

Put the two delta inequalities together to get $$c - \delta < x < c + \delta \Rightarrow |c - x| < \delta \left$$, which we know implies $$|f(x) - L| < \epsilon$$.

 

[scouter]

thnxxx mate.. I found it after a while... have a nice day...