Hi there,
The following is a suggested exercise I'm doing for my Analysis class.
Prove
Here's what I have, but it's straight from my class notes. I'm not sure what to do next.
f is continuous at a if:
Any ideas?
What do you mean by control? And where did the less than or equal to |x+a|delta come from? Sorry, I'm really having trouble with all of this. Switched from Social Sciences into Mathematics and Economics... the questions are a lot different than what I'm used to.
You shouldn't be "guessing" that, it is the definition of "limit"- besides it makes no sense until you have said what delta is! You want to have and you want to be able to guarentee that by choosing some so that "if then [tex]|x^2- a^2|< \epsilon[tex]".
As TheEmptySet told you, the crucial part is that |x+a|. You can always rewrite as and the only reason why you could not choose to be is that |x+a| is a function of x, not a constant.
So can you find some number, A, such that ? If so, then choosing will work:
If , then, working backwards, we would have .
And how do we find that A? We want . Assuming, for the moment, that A is positive, that is the same as . Well, of course, there is no A for which that is true for all x but we are talking about x being close to a since this is a limit as x goes to a. Lets require that the distance from x to a be no greater than a/2: |x- a|< a/2. (I am assuming that a> 0 here. I will leave the cases a< 0 and a= 0 to you.) That is the same as -a/2< x- a< a/2 so, adding 2a to each part, 2a- a/2= 3a/2< x+ a< 5a/2. If we choose A= 5a/2, that will be true and so we will have .
Normally, we are interested in the case that is very small number but there is no requirement that it be small. If we were just to choose then our requirement that |x-a|< a/2 might not be true. So what we do is choose to be the smaller of and .
Again, this is for a> 0. You do the cases a< 0 and a= 0.