Proving the One Sided Limit Theorem: A Rigorous Approach

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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 =/ ?
 
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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".
 
Is it possible to continue the solution? I can find the (1).
 
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.
 
thnxxx mate.. I found it after a while... have a nice day...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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