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Homework Help: A puzzling limit

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data

    limit (x->infinity) [pi/2-arctan x]^(1/x)

    2. Relevant equations

    3. The attempt at a solution

    The problem I am attempting to tackle is limit x->infinity [pi/2 - arctan x]^(1/x) ----- (1)

    I have attempted an array of solutions, most of which involve very messy derivatives involving L'Hospital's. I've tried taking the natural log of the equation and substituting in k for 1/x etc. After some research and a few hours of attempts the closest to a solution I got was the following:

    Using an inverse trig complimentary identity:
    arccot x = pi/2 - arctan x, Solving for pi/2 = arccot x + arctan x

    Substituting pi/2 for pi/2 into (1):
    limit x->infinity [(arccot x + arctan x) - arctan x]^(1/x)

    which yields:

    limit x->infinity [arccot x]^(1/x)

    from here i used the fact that cot(theta) = x/y. So if x->infinity in arccot x then this imply would mean y->0 and would be 0 on the unit circle.

    so we would hav: limit x->infinity [arccot x]^(1/x) = 0^0 which is an indeterminate form.

    However we could also consider 0 on the unit circle to be 2pi and then we would have:
    limit x->infinity [arccot x]^(1/x) = [2pi]^0 = 1. QED

    I feel in this attempt I am cheating the problem by abusing the unit circle, however all other attempts at manipulating 1/x*ln(pi/2-arctanx) didnt yield anything useful.
  2. jcsd
  3. Oct 22, 2013 #2
    Try the substitution ##y=\tan^{-1}x##. You might find ##\lim_{y\rightarrow\frac{\pi}{2}^-}(\frac{\pi}{2}-y)^\frac{1}{\tan y}## more manageable.
  4. Oct 22, 2013 #3
    Try taking the log of your expression giving you ##\frac {log(\pi/2) - arctanx}{x}##. What happens to arctanx as x ##\rightarrow \pi /2? ## Or what happens to tanx as x ##\rightarrow \pi /2?## and deduce from there.
  5. Oct 22, 2013 #4
    I doubt the validity of my answer, so it's not an infraction to post it, since it's basically a question.
    The limit of the given function multiplied by its inverse equals the product of their respective limits, which is, by definition, one. We know both limits are equal (this is the fishy part of my argument) because 1/x equals -1/x as x tends to infinite. So the answer is one.
    Last edited: Oct 22, 2013
  6. Oct 22, 2013 #5
    I tend to think it is not 1. See my post above.
  7. Oct 22, 2013 #6
    It's interesting to substitute 0^0 for (2*pi)^0 , but a function can have only one output value for a certain input value. It is however worth considering if this definition of a function has any fundamental significance, meaning concretely, if we for the sake of determining this limit supposed for a moment functions weren't defined this way, would this lead to an inconsistency?
    Last edited: Oct 22, 2013
  8. Oct 22, 2013 #7


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    The expression you posted ##\frac {log(\pi/2) - arctanx}{x}## doesn't come out correctly, it should be ##\frac {log(\pi/2 - arctanx)}{x}##. I think you should use l'Hopital on it and I think the limit of the original expression is 1.
  9. Oct 22, 2013 #8
    My method results in a correct answer, but is it a correct method?
  10. Oct 22, 2013 #9


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    I don't know. It doesn't sound right. Could you spell it out in more detail? If you are saying y*(1/y)=1 and the limits of y and 1/y are the same, yes, that gives you 1 (or -1). But how do you know the limits are the same? Sounds "fishy" to me.
    Last edited: Oct 22, 2013
  11. Oct 22, 2013 #10
    You are right. I mistyped the expression and indeed the final answer is 1. I forgot I had taken a log. Let's just paint me distracted. L'Hospital is fine.

    If you are ever unsure about L'Hospital you can back up to what it really means, which is comparing terms of the Taylor's polynomials of the top and bottom functions.

    Re your argument with inverses, it is a clever idea but you did not work it through correctly. The definition of the inverse ##f^{-1}(x)## is that ## f(f^{-1}(x))## = x. The product of their limits could be anything. If you're not sure, check out lim##_{x \rightarrow 2} x^2## and the same limit for its inverse ##\sqrt x##.

    Actually, by your argument, every function that has an inverse will have a limit of 1, regardless.

    If using inverses will help you out in a problem, fine, but please be sure there is an inverse before you invoke it. arctan x has an inverse only on an interval where tan x is increasing, say between ± ##\pi##/2. You can pick another interval if you wish, but there is no inverse over the entire x axis.

    It is perfectly true that ##lim_{x \rightarrow \infty}1/x## = ##lim_{x \rightarrow \infty}-1/x## = 0. They are approaching from different sides, but they will get there.
  12. Oct 23, 2013 #11
    The inverse terminology is a bit confusing, I was talking about the inverse power function.
    Essentially, how to prove:

    limit (x->infinity) [pi/2-arctan x]^(1/x) = limit (x->infinity) [pi/2-arctan x]^(-1/x)
  13. Oct 23, 2013 #12
    I solved the problem based off the suggestion from gopher when I had free time during my planning period and it worked out fine. This morning, I proved that it works for all pi/2*k- where k is an integer. Broke the proof down into two cases, where k=0 and k=/=0; the first involving the use of L'Hospitals and the later not.

    I'm not sure if you can get a result with out substitution: before when I just took the natural log of the original limit and applied L'Hospitals I kept getting indeterminate forms with no end. Gopher's suggestion lead to a much cleaner more direct result. Much thanks! If you all would like, I can attempt to submit the solution I arrived at.
    Last edited: Oct 23, 2013
  14. Oct 23, 2013 #13


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    Going from the log expression does require two applications of l'Hopital and some rearranging but it definitely ends and isn't THAT bad.
  15. Oct 23, 2013 #14
    I'll give it a go from that one then considering how I was doing this at the end of the day till 12 am, but I would love to see how you were able to work it out if you wouldn't mind!
  16. Oct 23, 2013 #15


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    I'll sketch it. Using l'Hopital on ##\frac {log(\pi/2 - arctanx)}{x}## gives you ##\frac {1}{\pi/2 - arctanx}## times other stuff. Put the other stuff in the numerator, check you have a 0/0 form then l'Hopital again, after that you have no more arctans and everything is powers of x. It's not hard from there.
    Last edited: Oct 23, 2013
  17. Oct 23, 2013 #16
    Hmm, I never thought taking the "stuff" into the numerator... I just distributed it to the bottom and boy did it get messy.
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