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Homework Help: How to find limit of sin^2(x)/x^2 as x--> infinity

  1. Dec 14, 2005 #1
    Hello Everyone,
    I'm doing some practice problems and I am stuck on the following limits.
    1) Limit as x approaches infinity of (sin^2(x))/x^2
    Now for this one I thought that the answer would just be 1 seeing as the limit of sinx/x is 1, however the answer is apparently 0 and I can't see the reasoning for this.
    2) Limit as x approaches 0 of (cosx-1)/xsinx
    Now for this one I first tried getting an the identity for sin^2x out of it by squaring it but that didn't work. Then I just tried breaking it up, but I just keep ending up getting 0 but apparently the answer is 1/2.
    3) Limit as x approaches pi/3 of (cosx-1/2)/(x-pi/3)
    Again I am totally lost. I keep wanting to try and get an identity out of it but I just cant get anything useful.
    4) Now this question I know should be easy. I am just having trouble understanding where to start. If F(x) = the integral of f(t)dt (the values on the integral are b=x and a=1) and f(t)= integral (squareroot(1+u^2))/u *du . Find F''(3).
    Apparently the answer to this is (2squareroot82)/3
    If this isn't clear I can always draw it out and post it, I'm sure that what I tryed to type isn't overly clear.
    Sorry about these questions guys, I know I haven't exactly shown a lot of work, but these are the ones that I have no idea of how to do, I've been attempting them for 2 days now and just can't seem to get anywhere :frown:
     
  2. jcsd
  3. Dec 14, 2005 #2

    Dale

    Staff: Mentor

    The limit of [tex]\frac{\sin (x)}{\Mfunction{x}}[/tex] is only 1 as x approaches 0, not as x approaches infinity. This problem is unrelated to that one.

    The key to this one is to realise that [tex]\frac{{\sin (x)}^2}{x^2}[/tex]is always between [tex]\frac{1}{x^2}[/tex] and 0.

    I think you can use L'hopital's (sp?) rule for 2 and 3.

    -Dale
     
    Last edited: Dec 14, 2005
  4. Dec 14, 2005 #3
    Hmmm ok, the way I interpreted that problem was in the form of (sinx)(sinx)/x^2 is this wrong? I always thought that sin^2x =(sinx)(sinx) and sinx^2/x^2 =[tex]\frac{{\sin (x)}^2}{x^2}[/tex]
     
  5. Dec 14, 2005 #4

    Dale

    Staff: Mentor

    Sorry about the confusion:
    [tex]{\sin (x)}^2 = \sin (x)\,\sin (x) =
    {\sin }^2 (x)[/tex]
    Mathematica just uses the notation that I used while most textbooks favor your notation.

    Either way, you are looking at the wrong part, in one case (sinx/x) you have the limit as x approaches 0, in the other (this problem) you have the limit as x approaches infinity. The squares have nothing to do with it in this case.

    -Dale
     
    Last edited: Dec 14, 2005
  6. Dec 14, 2005 #5
    Oh ok I see, my mistake I see what you are getting at now. I am still having trouble with the other questions though, I just don't see where to start :(
     
  7. Dec 14, 2005 #6

    Dale

    Staff: Mentor

    2 and 3 are just standard L'hopital's rule problems. Just apply his rule until you don't get 0/0.

    I didn't really follow 4. I might be able to help if you could TEX it, but I might still be lost.

    -Dale
     
  8. Dec 14, 2005 #7
    I will try and write it out and paste a link to number 4. As for 2 and 3 I dont actually know that rule so I will look it up on the net and see what I can come up with from that. Thanks
     
  9. Dec 14, 2005 #8
    Ok so with the L'Hopitals rule, all you have to do is take the derivative of the numerator and denominator and keep doing it until you don't get 0/0 anymore>???? Does this always work like this? Can I use this rule all the time in a situation like this? Based on reading about the rule this is how I solved the questions and I ended up with the right answer:

    2) cox-1/xsinx ---> -sinx/(sinx-xcosx)---> -cosx/cosx+cosx-xsinx which when you substitute in x=0 gives you an answer of -1/2!!!!

    3) (cosx-1/2)/(x-pi/3) -----> -sinx--->-sin(pi/3)---> -sqrt3/2!!!!

    If I can use that method all the time that is awesome! I never knew about it before, in class we were never shown that!!!

    Thanks a lot!!!!
     
  10. Dec 14, 2005 #9

    HallsofIvy

    User Avatar
    Science Advisor

    Do you know the "Fundamental Theorem of Calculus": If [tex]F(x)= \int_1^x f(t)dt[/tex] the F'(x)=f(x). In particular, [tex]F'(3)=f(3)= \int_1^{9} \frac{1+ u^2}{u}du[/itex].

    To do that integral, multiply both numerator and denominator by u to get
    [tex]F'(3)= f(3)= \int_1^{9} \frac{u(1+ u^2)}{u^2}du[/itex], and let v= 1+ u^2.
     
  11. Dec 14, 2005 #10

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    See you were never shown that in class, I would recommend to only use it as a test. Your teacher probably expects you to find it through another route.
     
  12. Dec 14, 2005 #11
    Where does the square root go in that integral? And how do you get the second derivative from that? Stupid questions I know, but I'm just now seeing this one .


    And how would you go about solving the limit problems without L'hopitals rule? Now that I learned it I rather like it....it's very handy!
     
  13. Dec 14, 2005 #12

    Dale

    Staff: Mentor

    There are (of course) some limitations. If you are trying to take the limit of some f(x)/g(x) then the most important limitation is that the limit of f'(x)/g'(x) must exist.

    For example, let [tex]f(x) = \sin (x)[/tex] and [tex]g(x) = x[/tex] then [tex]\Mfunction{Limit}(\frac{f(x)}{g(x)},
    {x\rightarrow {\infty}}) = 0[/tex]. However, [tex]f'(x) = \cos (x)[/tex] and [tex]g'(x) = 1[/tex] so [tex]\Mfunction{Limit}(\frac{f'(x)}{g'(x)},
    {x\rightarrow {\infty}})[/tex] does not exist. So for this one L'Hopital's rule doesn't work. Instead you have to use the same method as in problem 1 to get the right answer.

    -Dale
     
    Last edited: Dec 14, 2005
  14. Dec 15, 2005 #13

    VietDao29

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    Homework Helper

    ???
    What is this??? :confused:
    What do you mean???
    ---------------------
    L'Hopital rule is used when both numerator and denominator either tends to 0 or infinity, i.e:[tex]\frac{0}{0} \quad \mbox{or} \quad \frac{\infty}{\infty}[/tex]. It states that, if:
    [tex]\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)}[/tex] exists, and either:
    [tex]\lim_{x \rightarrow \alpha} f(x) = 0 \quad \mbox{and} \quad \lim_{x \rightarrow \alpha} g(x) = 0[/tex] or:
    [tex]\lim_{x \rightarrow \alpha} f(x) = \infty \quad \mbox{and} \quad \lim_{x \rightarrow \alpha} g(x) = \infty[/tex], then:
    [tex]\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}[/tex].
    -------------------
    #1, you cannot use L'Hopital rule since you don't know what sin2(x) tends to when x tends to infinity. It can be done using Squeeze theorem, by noticing the fact that:
    [tex]0 \leq \frac{\sin ^ 2 x}{x ^ 2} \leq \frac{1}{x ^ 2}[/tex]
    -------------------
    #2, you have two different ways to do it without using L'Hopital rule:
    The first way is to use power-reduction formulas:
    [tex]\sin ^ 2 x = \frac{1 - \cos(2x)}{2}[/tex].
    So here you have:
    [tex]\lim_{x \rightarrow 0} \frac{\cos x - 1}{x \sin x} = \lim_{x \rightarrow 0} - \frac{2 \sin ^ 2 \left( \frac{x}{2} \right)}{x \sin x} = \lim_{x \rightarrow 0} - \frac{2 \sin ^ 2 \left( \frac{x}{2} \right)}{4 \frac{x}{2} \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}[/tex].
    Can you go from here?
    The second way is to multiply both numerator and denominator by (1 + cos x), so that you'll have sin2x in the numerator.
    It goes like this:
    [tex]\lim_{x \rightarrow 0} \frac{\cos x - 1}{x \sin x} = \lim_{x \rightarrow 0} \frac{(\cos x - 1)(1 + \cos x)}{x \sin x (1 + \cos x)} = \lim_{x \rightarrow 0} \frac{- \sin ^ 2 x}{x \sin x (1 + \cos x)}[/tex].
    Can you go from here?
    -------------------
    #3, you can change 1 / 2 into: [tex]\frac{1}{2} = \cos \left( \frac{\pi}{3} \right)[/tex], then use the identity:
    [tex]\cos \alpha - \cos \beta = -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right)[/tex]. Can you go from here?
    L'Hopital rule is overkill... and as a matter of fact, you haven't learnt it, so you shouldn't use it.
    --------------------
    If [tex]F(x) := \int_{\alpha} ^ {x} f(t) dt[/tex] then [tex]F'(x) = \frac{d \left( \int_{\alpha} ^ {x} f(t) dt \right)}{dx} = f(x)[/tex]. There should be a proof in your book, or you can view it here (see part 3).
    So [tex]F''(x) = f'(x) = \frac{d \left( \int_{1} ^ {x ^ 2} f(u) du \right)}{dx} \ = \ ?[/tex], where [tex]f(u) := \frac{\sqrt{1 + u ^ 2}}{u}[/tex]. Note that the upper limit of the integral is x2, not x.
    Can you go from here?
     
    Last edited: Dec 15, 2005
  15. Dec 15, 2005 #14
    OK, thanks so much! I think I understand those limits now even without that handy little rule. And I can now get the answer for that integral! Thanks very much guys!
     
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