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One sided limit question.

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data

    If lim (x->0+) f(x) = A and lim (x->0-) f(x) = B

    Find lim(x->0+) f(x^3-x)

    2. Relevant equations



    3. The attempt at a solution

    I'm not sure how to do this. We know the right and left hand limits at x, how is it possible to find the right hand limit at x^3-x?
     
  2. jcsd
  3. Jun 24, 2009 #2

    Mark44

    Staff: Mentor

    If you haven't already done so, take a look at the graph of g(x) = x3 -x = x(x2 - 1). As x --> 0+, what does g(x) approach? As x --> 0-, what does g(x) approach?
     
  4. Jun 24, 2009 #3
    g(x) approaches 0 from both sides.
     
  5. Jun 24, 2009 #4

    Mark44

    Staff: Mentor

    As x --> 0+, g(x) --> 0+, right?
     
  6. Jun 24, 2009 #5
    Yes g(x) --> 0+

    I don't know what i'm missing :/
    That makes it lim (x->0+) f(0) ?
     
  7. Jun 24, 2009 #6

    Mark44

    Staff: Mentor

    You're trying to find
    [tex]\lim_{x \rightarrow 0^+} f(g(x))[/tex]

    As y -->0+, f(y) approaches which value, A or B?
     
  8. Jun 24, 2009 #7
    It approaches A?
     
  9. Jun 25, 2009 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Let u= x3- x. u is a polynomial in x and all polynomials are continuous so, as x goes to 0, u goes to 0. BUT if x= 0.001, x3= 0.000000001 so x3- x= 0.000000001-0.001= -0.000999999. u is NEGATIVE for x between 0 and 1. If x= -.001, x3= -0.000000001 so x3- x= -0.000000001+0.001= 0.000999999. u is POSITIVE for x< 0.
     
    Last edited: Jun 25, 2009
  10. Jun 25, 2009 #9
    Thank you! Now i get it :)
    So as x-->0+ f(x^3-x) has the limit B and as x-->0- it has the limit A.

    All i needed to do was take some numbers and see what happens..
    Is there a way to show this with symbols instead of words like this? Just wondering.
     
  11. Jun 25, 2009 #10

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, you could note that x3- x= x(x- 1). If x< 0 both of those are negative so the x3- x> 0 for all x< 0 and if 0< x< 1, x is positive while x- 1 is negative so x3-x< 0 for 0< x< 1.
     
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