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Homework Help: Please Check proof.

  1. Jan 2, 2012 #1
    Problem: Determine the limit l for the given a, and prove that it is the limit by showing how to find a δ such that abs(f(x) - l)< ε for all x satisfying 0< abs( x - a) < δ.

    ** abs = absolutue value of....

    f(x) = x(3-cos(x^2))

    The limit l is zero. hence the product of some number times zero is zero.

    abs(f(x) - 0)< ε = abs(x(3-cos(x^2)).

    abs((3-cos(x^2))≤ 1, abs(x(3-cos(x^2))< abs(x)

    Thus for every number x, If abs(x) < ε Then abs(x(3-cos(x^2)),

    Which finishes the proof, concluding that the limit of f(x) near zero is zero.
  2. jcsd
  3. Jan 2, 2012 #2


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    For future reference, use | for absolute value. It's easier to read.

    Is the problem to show that [itex]\lim_{x \to 0} f(x) = 0[/itex]?

    I'm not sure where this comes from.

    The bolded part is good start. I'm not sure why you're equating epsilon to |x(3-cos(x2)|.

    Neither of these statements are true. Do you see why?

    It's unclear how you concluded the bolded part.

    Then what? Is there supposed to be an equation here?
  4. Jan 2, 2012 #3
    "I'm not sure why you're equating epsilon to |x(3-cos(x2)|."
    Are you asking me why i said |x(3-cos(x2)| < ε?

    "Neither of these statements are true. Do you see why?"
    |x(3-cos(x2)| is not less than or equal to 1.. is that right?
    abs(x(3-cos(x^2))< abs(x), I am not sure why that is not correct...?

    "It's unclear how you concluded the bolded part."

    Because I said |x(3-cos(x2)| < ε, then I found out |x(3-cos(x2)| < |x|
    Then I assumed |x(3-cos(x2)| < |x - 0| < ε Thus |x| < ε
  5. Jan 2, 2012 #4


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    No. I'm asking why you're setting ε = |x(3-cos(x^2)|.


    -1 ≤ cos(x2) ≤ 1, so:

    |x(3-cos(x2)| ≥ |x(3-1)| = |2x| > |x|. This is the opposite of what you want.
    Last edited: Jan 2, 2012
  6. Jan 2, 2012 #5
    -1 ≤cos(x2) ≤ 1, Im assuming you picked +- 1 because of the range of the function?
    In general when proving limits how do I determine the appropriate interval for my function? For example If I had some polynomial g could I say -n ≤g(x) ≤ n, where n is an integer?
  7. Jan 2, 2012 #6


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    Correct. Cosine is bounded between -1 and +1.

    You don't want to use integers. That's what the epsilon is for. By letting ε > 0, this covers every case.
  8. Jan 2, 2012 #7
    The books solution was...
    -1 ≤ cos(x2) ≤ 1, so |(3-cos(x2)| <=4, and thus |x(3-cos(x2) - 0| = |x|*|x(3-cos(x)|<= 4|x|.

    So we take delta = ε/4.

    So in the first step they said -n ≤ cos(x2) ≤ n where n = 1, so If I had another situation like this example but the function was say a polynomial... then How would I pick n.
    I hope I am making sense.

    I understand that I have to use epsilon when im saying |f(x) - l|.
  9. Jan 2, 2012 #8


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    I know that gb7nash asked you this, but I didn't see any answer.

    Precisely, what limit are you trying to evaluate? Mostly, what is a ?

  10. Jan 2, 2012 #9
    a is zero, so the limit is zero..
    Sorry I forgot to put the value of a in my problem.
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