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Limit question

  1. Sep 9, 2016 #1
    (Mentor note: moved from another forum hence no template)

    For what value of the constant c is the following function continuous at x = −3?
    f(x)=(1/x+1/3)/(x+3) if x≠ -3
    f(x)=c if x=-3


    please provide proof... I am so confused:(
     
    Last edited by a moderator: Sep 9, 2016
  2. jcsd
  3. Sep 9, 2016 #2

    fresh_42

    Staff: Mentor

    Please provide the reason of your confusion ...
     
  4. Sep 9, 2016 #3
    I am confused because the second part of the function is just "c" which is a constant does not involve x.
     
  5. Sep 9, 2016 #4

    fresh_42

    Staff: Mentor

    What happens with ##f(x)## for ##x \neq -3## if you simplify the fractions? Have you tried to draw a graph around ##x = -3##?
     
  6. Sep 9, 2016 #5
    I have tried to simplfy the function, it is f(x)=1/(3x) and I noticed for the simplfied function, x can be -3
     
  7. Sep 9, 2016 #6
    oh, is C=1/(3(-3))=-1/9 ??
     
  8. Sep 9, 2016 #7

    fresh_42

    Staff: Mentor

    And what is ##f(-3)=c## in this version? By the way: ##x=0## is also forbidden in the original definition of ##x##.
     
  9. Sep 9, 2016 #8

    fresh_42

    Staff: Mentor

    Yes. So now you have to find an argument, why ##f(x)## becomes continuous if we set ##f(-3)= -\frac{1}{9}##.
     
  10. Sep 9, 2016 #9
    Many thanks for your help !!!
     
  11. Sep 9, 2016 #10
    On the oringinal function when x= -3, the function is undefined. However, if we set f(-3)=-1/9 then the oringinal function will become continous on x=-3.
    I think I understand now, thank you again for your help
     
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