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Limits and Continuity of a Piecewise Function

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Find a value for k to make f(x) continuous at 5

    f(x)= sqrt(x2-16)-3/(x-5) if x cannot equal 5
    3x+k when x=5

    2. Relevant equations
    none

    3. The attempt at a solution
    lim x->5 sqrt((x+4)(x-4))-3/(x-5) * sqrt((x+4)(x-4))+3/sqrt((x+4)(x-4))+3
    lim x->5 (x+4)(x-4)-9/(x-5)[sqrt((x+4)(x-4))-3]


    And that's as far as I got. I'm not sure what my next step should be or if what I did is wrong. I graphed the function and used a program to solve for the limit which is apparently 5/3, but I couldn't come up with that answer. I would really appreciate any help/suggestions. Thanks.
     
    Last edited: Oct 15, 2011
  2. jcsd
  3. Oct 15, 2011 #2

    vela

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    You need to use parentheses to make it clear what you're doing. It looks like what you are doing is finding
    [tex]\lim_{x \to 5} \frac{\sqrt{(x+4)(x-4)}-3}{x-5}\cdot \frac{\sqrt{(x+4)(x-4)}+3}{\sqrt{(x+4)(x-4)}+3}[/tex]
    My first question to you is, why? I don't see why you have the x-5 in there. Look up the definition of continuity. You should see you're making this problem harder than intended. Or did you not describe f(x) correctly in the beginning of your post?
     
  4. Oct 15, 2011 #3
    Sorry! I copied the question wrong the first time. I fixed it. It is all over (x-5) in the first part of the function. Sorry about that.
     
  5. Oct 15, 2011 #4

    SammyS

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    Compare limx→5f(x) amd f(5) .
     
  6. Oct 15, 2011 #5

    vela

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    Oh, okay. Then you're headed in the right direction. Like SammyS says, calculate the limit and compare it to f(5).
     
  7. Oct 15, 2011 #6
    I figured it out, thank you!
     
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