delta of epsilon

mikau · 2006-09-10 05:08:59

this is a topic that often seems to confuse me. I'm trying to get it clear in my mind and eliminate the little ghosts of uncertainty that are still lurking around in my mind. I'm going to explain how I'm understanding it so you guys can see if its right and make comments or pointers if you can.

Ok, a formal definition of a limit is:

The limit of f(x) as x approaches a is L when

if L - E < f(x) < L + E then a - d < x < a + d (where (x != a) )

Now one thing I think I just cleared up which I believe was a cause for confusion for a while with me, is this method does not allow you to FIND the limit, it merely allows you to CHECK or confirm that a limit is correct.

And d can usually (if not always) be found as a function of E. And E is how far f(x) can stray from L.

Now another thing that confused me for a while, if we want f(x) to be within E units of 5, and we find that a -d(E) < x < a + d(E) how does that prove the limit of f(x) as x approaches a is L? My explanation is it has to be within some limited distance of a. (here the distance is d) so if we allow x to become infinitely close to a, then whatever the minimal distance from a was needed, it will be overcome. Correct?

mikau · 2006-09-10 06:54:40

Oh man.. I am completely lost with this problem. I hate absolute value equations....

proove the limit of x^2 as x approaches 3 is 9. [ hint: Write | x^2 - 9| as |x - 3||x + 3|. Show that if | x - 3| < 1 then | x + 3 | < 7. If you let d (delta) by the smaller of the numbers 1 and E/7, show that this d works. ]

I'm completely clueless. That hint makes little sense to me.

mikau · 2006-09-10 09:23:26

come on, peoples! Anyone got a solution? :-(

Math Is Fun Forum

#1 2006-09-10 05:08:59

delta of epsilon

#2 2006-09-10 06:54:40

Re: delta of epsilon

#3 2006-09-10 09:23:26

Re: delta of epsilon

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