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Continuity and Differentiability

  1. Sep 1, 2013 #1
    1. The problem statement, all variables and given/known data

    f(x) = sin ∏x/(x - 1) + a for x ≤ 1
    f(x) = 2∏ for x = 1
    f(x) = 1 + cos ∏x/∏(1 - x)2 for x>1

    is continuous at x = 1. Find a and b
    2. Relevant equations

    For a lim x→0 sinx/x = 1.

    3. The attempt at a solution

    I tried equating the two limits, x →1+ and x→1- to each other and to 2∏. Didn't work.
    Maybe I something wrong, don't have any clue on how to proceed now. Can I get a clue?
     
  2. jcsd
  3. Sep 1, 2013 #2

    CompuChip

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    Homework Helper

    First let's get the notation straight. By sin ∏x/(x - 1) + a, do you mean $$\frac{\sin(\pi x)}{x - 1} + a?$$
    And then 1 + cos ∏x/∏(1 - x)2, is that $$1 + \frac{\cos(\pi x)}{\pi(1 - x)^2}$$ or $$1 + \frac{\cos(\pi x)}{\pi} (1 - x)^2?$$

    Anyway, you should be able to find the limit as ##x \to 1## for both parts separately. For example, try showing that
    $$\lim_{x \to 1} \frac{\sin \pi x}{x - 1} = - \pi.$$

    As you have already correctly remarked, continuity of f implies that
    $$\lim_{x \to 1} f(x) = f(1) = 2 \pi$$
    so you should then choose a and b such that these values are equal to ##2\pi##.
     
  4. Sep 1, 2013 #3
    Oops, sorry, it was [1 + cos∏x]/{[∏(1 - x)2}

    I put x = (1 - h) and as you said got lim x → 1- = a -∏
    and lim x → 1+ = ∏/2 + b.

    Thus, I got a = 3∏ and b = 3∏/2, which happen to be the right answer. Thank you!
     
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