Left and right hand limis.

I have been given the question:
Prove that the Limit of f(x) as x \rightarrow a is L if and only if the right hand limit of f(x)=the left hand limit of f(x)= L.

I have been able to prove that the right hand limit = L by assuming a a<x<a+\delta which gives that 0<|x-a|< \delta which implies |f(x)-L|= \epsilon from the definition of a limit.

However, I am stuck on proving the right handed limit, and was hoping someone could help!

(I can also go on to prove that the limit of f(x) = L if and only if the left hand limit is equal to the right hand is equal to L, i just keep gettin stuck on this small proof!)

Hope someone can help! Thanks!

Reply 1

DFranklin

ciara.m

I have been able to prove that the right hand limit ...
However, I am stuck on proving the right handed limit ...This makes no sense to me. What exactly are you stuck on?

Reply 2

ciara.m

Sorry, i'll try to explain better. I am doin a real analysis course at uni, and we have been set a few questions. one of them is to prove that
Lim (x tends to a) of f(x) = L if and only if the left hand limit = right hand limit = L.
My lecturer gave us a few steps to follow.
1) Prove that the right hand limit = L (using definitions)
2) Prove that the left hand limit = L (using definitions)
3) Prove that if right hand limit = left hand limit = L, then Limit (as x tends to a) = L.

I am stuck of step 2. I can't seem to prove that the left hand limit = L. This is what i need help with!

Reply 3

DFranklin

Can you not see how to change your proof of (1) into a proof of (2)? If not, post your proof of (1) and I'll explain.

Reply 4

ciara.m

Proof (1)
The definition states that for all epsilon > 0, there exists delta > 0 sucha that for all x, 0 < |x-a| , delta implies |f(x)-L| < epsilon.
Assume a < x < delta + a. This implies 0< x-a < delta which implies 0 < |x-a| < delta which, from the definition above, implies |f(x)-a|< epsilon. Therefore, we have verified the definition of Lim (x tends to a from above) = L.

When I try to apply the same logic to proof 2, i end up with -(delta) < x-a < 0. I wasn't sure if this means that 0< |x-a| < delta. If it does, then I cam prove it, but I didn't want to assume!

Reply 5

DFranklin

Yes it does. if (x-a) < 0 then |(x-a)| = -(x-a). You know -delta < x-a < 0, so multiply by -1, you know 0 < -(x-a) < delta. Since |x-a| = -(x-a) you then have 0 < |x-a| < delta.

(You don't have to spell it out in as much detail as I have done here, but you do need to know how to do these kinds of manipulations. (Which is only A-level maths, after all)).