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Homework Help: LCT Limit Comparison Test for Improper Integrals

  1. May 20, 2013 #1
    Learning about the Limit Comparison Test for Improper Integrals. I haven't gotten to any applications or actual problems yet. Just learning the theory so far, and have a question on the very beginning of it.

    1. The problem statement, all variables and given/known data

    f(x) ~ g(x) as x→a, then [itex]\frac{f(x)}{g(x)} = 1[/itex]
    (that is, f(x) is asymptotic to g(x) as x→a, then [itex]\frac{f(x)}{g(x)} = 1[/itex])


    sin(x) ~ x as x→0, because of the fact [itex]\frac{sin(x)}{x}=1[/itex]

    2. Relevant equations

    My question here is what order are the assumptions or logic being made?

    Is the idea to say (A):
    "Well, since we know the limit [itex]\frac{sin(x)}{x}=1[/itex], then we can conclude sin(x)~x"

    or is the idea to say (B):
    "Well, since we know sin(x)~x as x→0, then we can conclude the limit [itex]\frac{sin(x)}{x}=1[/itex]"

    3. The attempt at a solution

    The reason it doesn't make sense is because if x→0, then sin(x)=1 and x=0. So it is not possible to say [itex]\frac{1}{0}=1[/itex]

    Yet, if we first prove that [itex]\frac{sin(x)}{x}=1[/itex], then we can say that sin(x)~x.

    I am having trouble reconciling the fact that sin(x) is said to "behave" like x as x→0 while f(0) = sin(0) is a completely different number than g(0)= x.

    When you look at f(x) and g(x) separately first, it seems it can't be concluded that f(x)~g(x), but when we look first at [itex]\frac{f(x)}{g(x)}[/itex], then since their ratio is 1, we can conclude that f(x)~g(x), even though they are completely different functions when the numerator and denominator are evaluated separately.
  2. jcsd
  3. May 20, 2013 #2


    Staff: Mentor

    No, ##\lim_{x \to a} \frac{f(x)}{g(x)} = 1##
    The operative concept there is that the limit is 1. It does not say that f(x)/g(x) = 1.
    sin(x)/x ≠ 1, but ##\lim_{x \to 0} \frac{sin(x)}{x} = 1##
    I would say the latter.
    No. sin(0) = 0.
    sin(0) = 0

    If you look at the graphs of y = sin(x) and y = x close to x = 0, both graphs nearly coincide.
  4. May 20, 2013 #3
    Oh wow, its been so long since looking at trig or calculus that I was treating the x variable in sin(x) as an x axis point on the Cartesian instead of the angle. When the "x" or adjacent side of the triangle is 0 on the unit circle, then the ratio of opposite/hypotenuse is of course 1. That was the mistake I made in the latter part of the post.

    As for the beginning of the post, thank you for also correcting the theorem. I'm trying to learn it from a video of a university lecture and for many parts of the video, the camera man seems to scan past what the professor is writing on the board, so it takes a lot of rewind and replay to see and absorb it all.

    Best Regards,
    Last edited: May 20, 2013
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