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Limit of arctanx

  1. Mar 27, 2009 #1
    I just wanted to know why the limit of arctanx as x approaches infinity is [tex]\frac{\pi}{2}[/tex]. It doesn't make any sense to me.
  2. jcsd
  3. Mar 27, 2009 #2
    Also, when determining if an improper integral diverges. It always diverges if the limit is infinity? my text doesn't say it just gives two examples where this is the case so I wanted to check before I generalized.

  4. Mar 27, 2009 #3
    I'm merely a first year mathematics student who just recently covered this material so if anyone sees a mistake in my post, please correct me.

    The lowest upper bound of arctan x is pie/2, and the function of arctan x is always increasing since (arctan x)' = (1/(1+x^2)) is always positive. The lowest upper bound of a nondecreasing function is the limit as x approaches infinity, so your limit is pie/2.

    As for your second question, I'm quite sure that the integral diverges if it approaches infinity.
  5. Mar 27, 2009 #4
    By limit, I assume you mean one of the bounds of the improper integral. This is not true. The improper integral:
    [tex]\int_1^\infty \frac{1}{x^2} dx[/tex]
    is finite, even though one of the bounds is infinite.
    As for the previous question, note that arctangent is defined to be an inverse function for the tangent function, which is not one-to-one. They choose the part of the tangent function that lies in the open interval (-[itex]\frac{\pi}{2}[/itex], [itex]\frac{\pi}{2}[/itex]) to define the inverse function, which gives you the range for the inverse function arctangent. The behavior of the function, you can ascertain from the graph of x = tan(y) where y is taken to be in that interval, as it is the same as the graph of y = arctan(x).
    Last edited: Mar 27, 2009
  6. Mar 28, 2009 #5


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    Well, what should arctan(x) be in the first place?

    It is the angle (measured in radians), for which the tangent equals x!

    Now, remember:

    As the angle value approaches pi/2, the corresponding tangent value approaches infinity.

    Since the mapping from angles to tangent values is bijective, it follows that we can define an inverse mapping (i.e, the arctan-mapping), having in particular, the property that as the tangent value approaches infinity, the angle value approaches pi/2.
  7. Mar 28, 2009 #6
    Just a little nitpick: the mapping is not bijective, but you can restrict it to a domain that will give a bijection with R.
  8. Mar 28, 2009 #7


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    Since you had already eminently mentioned the necessary domain restriction, I didn't bother to belabor that point. :smile:
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