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Limit of an inverse tangent function

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    limit as x->[tex]\infty[/tex] of arctan[(1/x)-1]

    2. The attempt at a solution

    Well, my graphing calculator is showing a vertical asymptote at x=0 and a horizontal one at x=(what I think is) -[tex]\pi[/tex]/4

    I'm not really sure how to solve this though... hence me being here : /

    EDIT: Nevermind, I figured it out by splitting it up into lim(arctan(1/x)) and lim(arctan(1)), which did indeed give me -pi/4. I'm not sure how to delete this though, so I'll just leave it up for anyone who's having a similar problem, I suppose. :)
     
    Last edited: Sep 14, 2009
  2. jcsd
  3. Sep 14, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    You got the right answer, but you did it wrong. arctan(1/x-1) is not equal to arctan(1/x)-arctan(1). You can't break it up that way. If you know lim(1/x)=0 then lim(1/x-1)=(-1). Then since arctan is continuous, lim(arctan(1/x-1))=arctan(lim(1/x-1))=arctan(-1).
     
  4. Sep 14, 2009 #3
    Ok take y=(1/x)-1. What y goes to as x goes to infinity?
     
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