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Solving tricky limit (involving defined integral and sine)

  1. May 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate ##\lim_{x\to 1} \frac{1}{x-1} \int_{1}^{f(x)} sin(\pi t^2) dt##. f is differentiable in the neighbourhood of point ##x=1## and ##f(1)=1##.

    2. Relevant equations
    If ##f## is continuous on a closed interval ##[a,b]##, then there exists ##ξ∈]a,b[## such that

    ##\int_{a}^{b} f(x)dx=f(ξ)(b-a)##.
    ( First Mean Value Theorem for Integrals)

    3. The attempt at a solution
    I think the trick is to use the First Mean Value Theorem for Integrals. Obviously ##\sin(\pi t^2)## is continuous everywhere and so for a fixed ##x≠1##; ##\frac{1}{x-1} \int_{1}^{x} sin(\pi t^2) dt=\frac{1}{x-1}(x-1)\sin(\pi ξ^2)=\sin(\pi ξ^2)##. Therefore ##\lim_{x\to 1} \frac{1}{x-1} \int_{1}^{x} sin(\pi t^2) dt=\lim_{x\to 1} sin(\pi ξ^2)=0 ##
     
    Last edited: May 17, 2016
  2. jcsd
  3. May 17, 2016 #2

    Samy_A

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    The hint is right there: "I am supposed to use the First Mean Value Theorem for Integrals".

    Apply this to ##\displaystyle \frac{1}{x-1} \int_{1}^{f(x)} \sin(\pi t^2) dt## for a fixed ##x \neq 1##, and remember that ##ξ∈]1,f(x)[## (or ##ξ∈]f(x),1[## to be precise).

    Then take the limit.
     
  4. May 17, 2016 #3
    Sounds simple, but not sure how to put it together. I edited my attempt at a solution in the 1st post.
     
  5. May 17, 2016 #4

    Samy_A

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    You lost the ##f(x)## in the upper integration limit.
    What you get is ##\displaystyle \frac{1}{x-1} \int_{1}^{f(x)} sin(\pi t^2) dt=\frac{1}{x-1}(f(x)-1)\sin(\pi ξ^2)##.
    Now take the limit for x→0.
     
  6. May 17, 2016 #5
    i applied LH Rule and applied newton leibniz rule for differentiation under integral sign

    and got :- ## \lim_{x \rightarrow 1} f'(x).sin( \pi (f(x))^2) ##

    is f differentiable at x=1?
    and how can solve further if we dont know ## f'(1) ##?? :sorry:
     
  7. May 17, 2016 #6

    Samy_A

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    That f is differentiable at x=1 is given, but nothing is stated about f' being continuous at x=1. So you can't just assume that ##\displaystyle \lim_{x \rightarrow 1} f'(x) =f'(1)## (whatever the value of ##f'(1)## is).
     
  8. May 17, 2016 #7
    yes, then how can we solve the question now?
     
  9. May 17, 2016 #8

    Samy_A

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    See post #4.

    Notice that it is given that ##f(1) = 1## and that ##f'(1)## exists.
     
  10. May 17, 2016 #9
    what i think is that since f(x) is differentiable near x=1 then f'(1+) and f'(1-) should have finite values and that gives me answer =0 ??
     
  11. May 17, 2016 #10

    Samy_A

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    Not sure.
    Take ##f: \mathbb R \to \mathbb R## defined by ##\forall x\neq 0: f(x)=x²\sin (1/x)##, ##f(0)=0##.
    Then ##f## is differentiable everywhere, but f'(0+) and f'(0-) do not exist.

    By the way, remember that we are not supposed to post solutions before OP has solved the exercise, just hints. :oldwink:
     
  12. May 17, 2016 #11
    k, next time i will remember :smile:
     
  13. May 17, 2016 #12
    I think i have a problem,
    i used first principle ( basic definition of LHD and RHD)
    ##f: \mathbb R \to \mathbb R## defined by ##\forall x\neq 0: f(x)=x²\sin (1/x)##, ##f(0)=0##.
    i get f'(0+) = f'(0-)=0

    but the derivative way fails here.??

    ## LHD~ at ~x=0 = \lim_{h \rightarrow 0+} \frac{(0-h)^2.sin(\frac{1}{0-h})-0}{-h} = \lim_{h \rightarrow 0+} h.sin(\frac {1}{h}) =0 ##

    ## RHD ~at ~x=0 = \lim_{h \rightarrow 0+} \frac{(0+h)^2.sin(\frac{1}{0+h})-0}{+h} = \lim_{h \rightarrow 0+} h.sin(\frac {1}{h}) =0 ##

    ## then~ f'(0+)=f'(0-)=0 ~hence,~ f~ is~ differentiable ~at x=0 ##
     
    Last edited: May 17, 2016
  14. May 17, 2016 #13

    Samy_A

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    Ok, we were talking at cross purposes here.
    I thought that by ##f'(1+)## you meant ##\displaystyle \lim_{x \to 1,x>1} f'(x)##.
     
  15. May 17, 2016 #14
    when i take derivative and directly put x=0 i get ## f'(0) = -cos(\infty) ## which is not defined , why does directly differentiation not work here but the basic definition does? o_O
     
  16. May 17, 2016 #15

    Samy_A

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    Because the derivative is not continuous at x=0.
     
  17. May 17, 2016 #16
    why is it not continuous at x=0, by post #12 it very much is continuous at x=0

    here is the graph:-

    img1.jpg
    zooming in near x=0
    img_2.jpg
     
  18. May 17, 2016 #17

    Samy_A

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    The derivative is not continuous at x=0.
    For x≠0, ##f'(x)=2x\sin(1/x)-\cos(1/x)##.
    If you take the limit for x→0, the fist term goes to 0, but the second term, ##\cos(1/x)##, oscillates between -1 and +1.

    EDIT: we better end this digression.
     
    Last edited: May 17, 2016
  19. May 17, 2016 #18

    pasmith

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    L'hopital is not that helpful here as you don't know whether [itex]\lim_{x \to 1} f'(x)[/itex] even exists. But even so, it might be that [itex]\lim_{x \to 1} f'(x)\sin( \pi f(x)^2)[/itex] exists, although proving it may require rather more work than the OP's mean value theorem approach.

    In the OP's initial post they applied the mean value theorem incorrectly, allowing them to simply cancel [itex](x-1)[/itex] from numerator and denominator. However applying the mean value theorem correctly as in post #4 requires one instead to show that [tex]\lim_{x \to 1} \frac{f(x) - 1}{x- 1}[/tex] exists.

    Also the OP needs to give more justification for their conclusion that [itex]\lim_{x \to 0} \sin(\pi \xi(x)^2) = 0[/itex].
     
  20. May 18, 2016 #19
    That's the definition of ##f'(1)## because ##f(1)=1##. The limit [tex]\lim_{x \to 1} \frac{f(x) - 1}{x- 1}=\lim_{x \to 1} \frac{f(x) - f(1)}{x- 1}[/tex] exists since ##f## is known to be differentiable at ##x=1##.

    edit: oops, in the neighbourhood of ##x=1##
     
    Last edited: May 18, 2016
  21. May 19, 2016 #20
    Yes,Thanks I realized that if f(x) is differentiable at x=a then it does not imply that f'(x) is continuous at x=a.
     
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