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For the love of God somebody save me!

  1. Feb 28, 2007 #1
    Ok, so it's my first semester of Calculus and I'm completely lost. My professor is this old Korean guy who doesn't speak English and I'm in desperate need of some understanding.

    1. |X| = {x if x >= 0 and -x if x <0
    lim x->a |X| Does not exists.

    My first question is can the limit of an absolute value function ever exists? I understand the mechanics, just not the concept :(.

    2. The Squeeze Theorem
    If f(X) <= h(X) <= h(x) when x is near a (except possibly at a) and

    lim x->a f(X) = lim x->A h(X) = L

    then

    lim x->a g(X) = L

    Question: Prove that lim x->0+ suareroot(X)[1 + sin^2(2PI/X)] = 0.

    Sorry for my semi-broken representation. It's a mixture of C++ sytanx and algebra... I don't think it works so well but I hope it can be understood. Anyway, I don't even know where to begin setting up the proof...

    Anyone have an tips that might help? I'm going to search the net and try and find something that can help.

    Shwank
     
  2. jcsd
  3. Feb 28, 2007 #2
    First, you should probably post a question like this in the homework section, since it will get more responses there, and also that is where homework is supposed to be posted (since I am guessing it is homework, or at least feels like it).

    For question 1 I am not sure what you are trying to do.

    |x| is continuous and thus the [itex]\lim_{x\to a}|x| = |a|[/itex]. So I guess to answer your question, yes the limit of the absolute value function can exist, and it exist for all [itex]x \in \mathbb{R}[/itex].

    For the second question you want to find a function that is below suareroot(X)[1 + sin^2(2PI/X)] and one that is above. Of course you want each of these functions such that the limit as x->0+ is 0. So try to think of some functions (ones that you can easily compute the lim as x-> 0 for) and verify that the lower function is below suareroot(X)[1 + sin^2(2PI/X)] and that the upper function is above it.

    Could you take the lower function as the zero function? That is, is 0 <= suareroot(X)[1 + sin^2(2PI/X)] for all x? And does the lim x>0+ of 0 = 0? Now try to find a function that is above, and show that it is above for all x, and that the limit x->0+ of that function is 0.
     
    Last edited: Feb 28, 2007
  4. Feb 28, 2007 #3
    Think of these few examples:

    Does the limit of |x^2| as x approaches 0 exist?
    Does the limit of |x| as x approaches 1 exist?

    Let f(x) = sin(x) if x>0, -sin(x) if x<0, and 0 if sin(x) = 0

    Does the limit of |f(x)| as x approaches 0 exist?

    Can you think of a function g(x), where g(x) is greater than x^(1/2)[1 + sin^2(2PI/X)] for all x and as x approaches 0 from the right g(x) tends towards 0?

    Can you think of a function h(x), where h(x) is less than x^(1/2)[1 + sin^2(2PI/X)] for all x and as x approaches 0 from the right h(x) tends towards 0?

    **hint** what is the maximal and minimal value of sin^2(2PI/X)
     
  5. Mar 1, 2007 #4
    Thank you both for the help. :)... With your hints, and help from tutors at school I was able to understand.
     
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