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Homework Help: Finding constant in a function

  1. Aug 20, 2010 #1
    Determine all values of the constant a such that the following function is continuous for all real numbers.

    f(x) = ax/tan x, if x≥ 0
    a^2 -2, if x<0


    I've tried to plug in zero but the tan x throws me off because plugging in zero will give me a 0/0. Other than that I am stuck on what to do.
     
  2. jcsd
  3. Aug 20, 2010 #2

    CompuChip

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    So the limit from the left as you approach zero is clearly a2 - 2.
    For the limit from the right, you get 0/0, which is an indeterminate form. Therefore, you can use for example L'Hopitals rule to try and find the limit.
     
  4. Aug 20, 2010 #3
    Hi can you explain the L'Hopitals rule? I'm sorry I'm just starting Calculus and my teacher assigned this problem and we have yet to learn it. Thanks.
     
  5. Aug 20, 2010 #4

    hunt_mat

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    To avoid L'Hopital's rule, note that:
    [tex]
    \tan x=x+\frac{x^{3}}{3}+\cdots
    [/tex]
    so:
    [tex]
    \frac{ax}{\tan x}=\frac{a}{1+x^{2}/3+\cdots}
    [/tex]
    Then the limit as x tends to zero is equal to a, so from here, can you answer the rest of the question?

    Mat
     
  6. Aug 20, 2010 #5
    Yes! Thank you for you help!
     
  7. Aug 20, 2010 #6

    HallsofIvy

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    I would be very surprized if a class had covered power series expansions of functions but not L'Hopital's rule!
     
  8. Aug 20, 2010 #7

    hunt_mat

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    If you can't do L'Hopital's rule or power series, how can you evaluate the limit? I suppose that if you were shown that:
    [tex]
    \lim_{x\rightarrow 0}\frac{\sin x}{x}=1
    [/tex]
    by some other method, and you know about the algrebra of limits, then you could evaluate by brute force I suppose.

    Mat
     
  9. Aug 20, 2010 #8

    Char. Limit

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    You're not going to find a constant that makes this continuous for all real numbers. The cotangent function is discontinuous at an infinite number of points, and multiplying it by x won't solve that problem.
     
  10. Aug 20, 2010 #9

    hunt_mat

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    I think he means continuous at the point x=0...
     
  11. Aug 21, 2010 #10

    HallsofIvy

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    That was my thought.
    [tex]\frac{ax}{tan(x)}= a\frac{x}{sin(x)}cos(x)[/tex]
    Since both cos(x) and x/sin(x) go to 1 as x goes to 0, this has limit "a". [itex]a^2- 2[/itex] is a constant so to have this function continuous at x= 0, we must have [itex]a^2- 2= a[/itex] which is equivalent to [itex]a^2- a- 2= (a- 2)(a+ 1)= 0[/itex]. That has the two roots a= 2 and a= -1.

    However, I note that the original post specifically said "continuous for all real numbers" and, as Char. Limit said, no matter what a is, the function is not continuous for x any multiple of [itex]2\pi[/itex] except 0.
     
  12. Aug 22, 2010 #11

    hunt_mat

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    Incorrect posting perhaps?
     
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