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Continuity & Discontinuity in limits

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    a) Determine the points where the function f (x) = (x + 3) / (x^2 − 3x − 10) is discontinuous. Then define a new function g that is a a continuous extension of f .

    b) Determine what value of the constant k makes the Piecewise function
    { (x − k)/ (k^2 + 1) , x < 0
    f (x) = { x^2 + k , x ≥ 0

    continuous on lR. Justify your answer.

    c) True or False: if f and g are both continuous at 0, then f ◦ g is also continuous at 0. Justify your answer by proving it (if true) or giving a counterexample (if false).


    2. Relevant equations



    3. The attempt at a solution

    a) it is discontinuous at 5 and -2.Factored the denominator.

    b) lim x-> 0- = k
    & lim x-> 0+ = k

    c) lim x-> 0 f(x) = f(0) & lim x-> 0 g(x) = g(0) so, f ◦ g is lim x-> 0 f(g(x)) = 0. It is continuous.

    I hope I get a reply..No one is helping on other sites either. Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: Sep 29, 2011
  2. jcsd
  3. Sep 29, 2011 #2
    a) Fix the discontinous points, give g any value at those points

    b)

    You have

    [tex]f(x) = \begin{cases}
    \frac{x-k}{k^2 + 1} &\text{if } x < 0\\
    x^2 + k &\text{if } x \geq0
    \end{cases}[/tex]

    And you said that

    [tex]\lim_{x\to0^-} \frac{x-k}{k^2 + 1} = \lim_{x\to0^+}x^2 + k[/tex], find that k

    c) That seems right.
     
  4. Sep 29, 2011 #3
    I still don't understand a) . I got b). How sure are you about c) ?
     
  5. Sep 29, 2011 #4

    HallsofIvy

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    This is wrong. Try again.

    No, you do not know that the limit is 0 nor do you know that you should be looking at limit of g(x) as x goes to 0 because you do not know that f(0)= 0.

     
  6. Sep 29, 2011 #5
    what is wrong ? Give me an explanation or something so i understand better.
     
  7. Sep 29, 2011 #6
    Nope...I read this as what HallsofIvy pointed out. It says the f and g are continous at x = 0, but f and g may not be 0.
     
  8. Sep 29, 2011 #7

    HallsofIvy

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    Well, how about you tell how you got that
    [tex]\lim_{x\to 0}\frac{x- k}{k^2+ 1}= -k[/tex]!
    What do you get if you just take x= 0?
     
  9. Sep 29, 2011 #8
    Explanation..please
     
  10. Sep 29, 2011 #9
    ok

    -k = k
    -k - k = 0
    -2k = 0
    k = 0/-2
    k = 0

    Explain the other ones please
     
  11. Sep 29, 2011 #10
    Guys...Please help me.
     
  12. Sep 29, 2011 #11
    What is 0 - k?
     
  13. Sep 29, 2011 #12
    -k ....what kinda question is that LOL. I know that.
     
  14. Sep 29, 2011 #13
    Please help me quickly..it is getting late.
     
  15. Sep 29, 2011 #14
    Can you guys please explain it to me a) and c) because i already got b) ..
     
  16. Sep 29, 2011 #15
    [tex]\frac{0-k}{k^2 + 1}[/tex]
     
  17. Sep 29, 2011 #16

    SammyS

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    What if g(0) = a, where a ≠ 0.

    What do you know about f(a) ?
     
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