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Find the value of a and b for which the function is continuous at 2

  1. Sep 23, 2014 #1
    1. The problem statement, all variables and given/known data

    [tex]
    u(x) =
    \begin{cases}
    \frac{3x+b}{4} & \text{if } x \geq 2 \\
    \frac{(3-x)^n-a}{x-2} & \text{if } x < 2
    \end{cases}
    [/tex]
    find the value of a and b for which the function is continuous at 2

    3. The attempt at a solution
    I tried to proof that lim(3x+b)/4 = lim (3-x)^n-a/x-2 = f(2)
    that gives lim (3-x)^n-a/x-2= 6+b/4
    But I have a problem with the limit when x< 2, I don't know how to solve it
    Can u please help ?
     
  2. jcsd
  3. Sep 23, 2014 #2

    HallsofIvy

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    The difficulty with x< 2 is, of course, that the denominator is 0 at x= 2. In order for that to have a chance at having a limit, the numerator must also be 0 at x= 2. That is, we must have [tex](3- 2)^n- a[/tex]. What does that tell you about a?
     
  4. Sep 23, 2014 #3
    It means that a=1, and after using the formula for A^n-B^n the limit when x< 2 becomes -(2^n-1+2^n-2+....+1) which is -2^(n-2) and then I can find b which is -2^(n-1)/3
    can u tell me if that is correct ? and should I also find f(2)?
     
  5. Sep 23, 2014 #4

    haruspex

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    I don't get the same answer for that limit. Please post your detailed working.
     
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