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Homework Help: L'Hospital's Rule and Integration

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data

    lim (x→0) of the definite integral (from 1 to 1+x) of (cos(t))/t all over x

    2. Relevant equations
    Integration by parts
    L'hospital's rule

    3. The attempt at a solution
    I believe that I must first solve the definite integral, and then take this result and use L'hospital's rule in order to solve the equation. However, using integration by parts on the integral has lead me nowhere. I was hoping somebody could point me in the correct direction. Thanks.
  2. jcsd
  3. Apr 2, 2014 #2

    Ray Vickson

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    Homework Helper

    You are mis-using l'Hospital's rule. Go back and review it; you will soon see what is happening.
  4. Apr 2, 2014 #3
    I know how to use the rule and I just assumed that it would be needed here, the real problem I am having is with the definite integral. Can I just write it in terms of 'x' using The Second Fundamental Theorem?
  5. Apr 2, 2014 #4
    Never mind. The reply below has the better approach.
    Last edited: Apr 2, 2014
  6. Apr 2, 2014 #5
    Given ##F(x)=\int_1^{1+x}\frac{\cos t}{t}\ dt##, you have correctly identified that ##\lim_{x\rightarrow 0}F(x)=0## and that l'Hopital's rule might prove useful in evaluating ##\lim_{x\rightarrow 0}\frac{F(x)}{x}##.

    Might it be easier to use the First Fundamental Theorem of Calculus to find ##F'(x)## than what you have tried?
  7. Apr 2, 2014 #6
    Would there be a way to do it without using the Taylor series? Technically in my class we have not learned it yet so I believe it is supposed to be solved without the use of a series.
  8. Apr 2, 2014 #7
    I can't use the First Theorem because I do not know how to integrate the function
  9. Apr 2, 2014 #8
    Recall that the theorem states that "If f is continuous on the interval [a, b], then the function defined by [itex]F(x) = \int_a^x f(t) dt[/itex] is differentiable on (a, b) with derivative F'(x) = f(x)." In other words, you do not need to know an antiderivative of f(t). You just need to know whether it is continuous or not. :)
  10. Apr 2, 2014 #9


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    Use the Second Theorem.
  11. Apr 2, 2014 #10
    So does this mean I could just write the problem all in terms of x and then apply L'Hopsital's theorem? (ie, differentiate the top and the bottom of the equation?)
  12. Apr 2, 2014 #11
    Well, to be sure, what are you going to write as the derivative of the numerator [tex]\int_1^{1+x} \frac{\cos t}{t}\, dt \; ?[/tex] It will indeed be a function of x, as the variable t is just a dummy variable of integration.
  13. Apr 2, 2014 #12
    In most texts that I have worked with, the First Fundamental Theorem essentially states that, for ##F(x)=\int_a^xf(t)\ dt##, ##F'(x)=f(x)##. I usually tell my students that it's the easiest derivative formula they're ever gonna come across. It allows one to find the derivative of a so-called "accumulation function" with almost no effort whatsoever.

    A more general version of the First Fundamental Theorem states $$\frac{d}{dx}\left(\int_{u(x)}^{v(x)}f(t)\ dt\right)=f\big(v(x)\big)v'(x)-f\big(u(x)\big)u'(x)$$ whcih is derived from the "normal" First Fundamental Theorem, properties of definite integrals, and the chain rule for derivatives. This is the version that applies here.
  14. Apr 2, 2014 #13
    I was going to write (1+x)(Sin(1+x)) + (Cos(1+x))/ (1+x)^2
  15. Apr 2, 2014 #14
    After reading about the Second theorem I am thinking it may just be Cos(1+x)/(1+x)
  16. Apr 2, 2014 #15
    I see. That's not quite correct. Let me apply the First Fundamental Theorem to a different function so you can see how it is used. Suppose we want to know the derivative of [tex]g(x) = \int_1^{x + 3} \sin(t)\, dt[/tex]
    Since sin(t) is continuous on any interval [1, b], the First Fundamental Theorem tells us that the function [itex]F(x) = \int_1^x \sin(t)\, dt[/itex] is differentiable on (1, b) with derivative F'(x) = sin(x). Until you gain more experience with applying it, always use the exact wording of the theorem.
    Our function is therefore a composition: g(x) = F(x + 3). Therefore, the chain rule tells us that g'(x) = F'(x + 3) * (x + 3)' = sin(x + 3)*1 = sin(x + 3). Does that help ?
  17. Apr 2, 2014 #16
    There you go! :)
  18. Apr 2, 2014 #17
    It does help, now I just need to apply L'hospital's rule, correct?
  19. Apr 2, 2014 #18
    Ok, so I don't think I need L'hospital's, applying the limit I get cos(1), correct?===> Nevermind, I don't think this answer is correct
  20. Apr 2, 2014 #19
    Remember that we are finding the derivative of [itex]g(x) = \int_1^{1+x} \frac{\cos t}{t}\, dt[/itex], which is an integral, not a quotient. The First Fundamental Theorem never requires us to find the derivative of the function inside the integral. It tells us that the derivative is the function inside the integral. That is, we are definitively not find the derivative of [itex]\frac{\cos t}{t}[/itex], as that quantity is not even a function of x! Only the integral varies with x, and the amount that it varies is precisely the value of the function being integrated.
    This is fundamentally connected to what we do when we integrate: we add up the values of the function over each interval scaled by the size of each interval as that size approaches 0. So it makes sense that the rate at which the integral varies at each point is the same as the value of the function being added at that point.
    Edit: If you meant L'Hopital's rule, then no, you do not differentiate the quotient. As long as the numerator and denominator individually approach 0, you differentiate them as separate functions.
    Last edited: Apr 2, 2014
  21. Apr 2, 2014 #20
    Thank you, so I took the limit as n->0 of ((Cos(1+x)/(1+x))/x and found that the limit does not exist
  22. Apr 2, 2014 #21
    However, I simplified the equation and found the limit to be zero, which was not one of the answer provided, so I think I may have made a mistake evaluating the limit
  23. Apr 2, 2014 #22
    You differentiated the numerator, which was the integral, but you didn't differentiate the denominator of x. Your new limit, due to applying L'Hopital's rule to the original limit, should be
    [tex]\lim_{x\rightarrow 0} \frac{\int_1^{1+x} \frac{\cos t}{t}\, dt}{x} = \lim_{x\rightarrow 0} \frac{\frac{\cos(1 + x)}{1+x}}{1}[/tex]
    That definitely has a real limit. :)
  24. Apr 2, 2014 #23
    I see, thank you very much!
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