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Right limit of a composite function. Original functions limits are known.

  1. Dec 21, 2011 #1
    1. The problem statement, all variables and given/known data

    If [tex]\lim_{x\rightarrow 0+}f(x)=A [/tex]
    and
    [tex]\lim_{x\rightarrow 0-}f(x)=B[/tex] find

    [tex]\lim_{x\rightarrow 0+}f(x^{3}-x)[/tex]


    2. Relevant equations



    3. The attempt at a solution

    I dont have one. Im dumbfounded. Mostly i have been trying to understand the meaning of the composite function. Im not sure if this is correct but

    [tex]f(x^{3}-x) =f(x(x^{2}-1))= f(x)f(x^{2}-1)[/tex]

    Then i have at least isolated f(x), whose limit is known, but the other factor i dont know what to do about.
    I've been thinking about the meaning of the limits for the original function f(x). Since A is not equal to B, f(x) is not even. f(x) could be odd, but we dont know that. I dont know how that would help me, its just something i thought of.
     
    Last edited: Dec 21, 2011
  2. jcsd
  3. Dec 22, 2011 #2

    Mark44

    Staff: Mentor

    No, it's not correct at all. It is not generally true that f(ab) = f(a)f(b).

    Under certain conditions, though, you can reverse the order of taking a limit and evaluating a function. Your textbook probably has a theorem that deals with this.
     
  4. Dec 22, 2011 #3

    SammyS

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    What is the sign of (x3-x), when 0 < x < 1 ?
     
  5. Dec 22, 2011 #4
    Oh ok, darn. I just tested it with f(x) = Ax + B, wich was stupid.
    I was trying to find a theorem about it in my textbook (Calculus by Adams) but failed. Any tips on books that can teach me more about composite functions?

    Negative, since x3 < x there.
     
  6. Dec 22, 2011 #5
    Am i wrong in assuming that if for instance

    [tex]f(x)=Ax+B[/tex]

    then

    [tex]f(x^{3}-x)=(Ax+B)^{3}-(Ax+B)[/tex] ?
     
  7. Dec 22, 2011 #6

    HallsofIvy

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    Yes, you are wrong. You have the order of composition backwards. If [itex]f(x)= x^3- x[/itex] then [itex]f(Ax+ B)= (Ax+B)^3- (Ax+B)[/itex].

    With f(x)= Ax+B, then [itex]f(x^3- x)= A(x^3- x)+ B[/itex].

    In any case, this has nothing to do with the original question. You know That [itex]x^3- x< 0[/itex] for x between 0 and 1 and, of course, goes to 0 as x goes to 0. And you know that [itex]\lim_{x\to 0^-}f(x)= B[/itex].
     
  8. Dec 23, 2011 #7
    Oh right, thats how compositions are structured! Darn i have forgotten alot.

    Ok, so since we have x^3 - x in place of x, in the original function f(x), and x^3 - x < 0 when 0 < x < 1, we have created the situation

    [tex]\lim_{x\rightarrow 0-}f(x)[/tex]

    Wich is B. The answer is B.

    Did i understand it now?
     
    Last edited: Dec 23, 2011
  9. Dec 23, 2011 #8

    gb7nash

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    That's correct.

    What about the case [itex]\lim_{x \to 0^+} f(x^3+x)[/itex]?
     
  10. Dec 23, 2011 #9
    Ok, since [tex] x^{3}+x > 0[/tex] when [tex] 0 < x < 1[/tex] and since [tex]x^{3}+x \rightarrow 0[/tex] when [tex] x \rightarrow 0[/tex] we should get

    [tex]\lim_{x\rightarrow 0+}f(x^{3}+x) = A[/tex]

    Am i right?
     
  11. Dec 23, 2011 #10

    Dick

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    I'd agree with that.
     
  12. Dec 23, 2011 #11
    Thank you very much Mark44, SammyS, HallsofIvy, gb7nash and Dick.
     
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