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Finding a limit using substitution rule, my answer is 0, my book`s is -2

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Lim [(tanx)^2] / [1 + secx] <<< as x goes to pi



    2. Relevant equations



    3. The attempt at a solution

    (tan x)^2 = (sin x)^2 / (cos x)^2

    (sin x)^2 = y

    lim y = 0 <<< as x goes to pi


    lim [y/ (cos x)^2] / [1 + (1/y)] <<< as y goes to C=0

    1+ (1/y) = (y+1)/y

    lim [y^2] / (y+1) (cos x)^2

    y=0

    so, 0/(0+1)(cos 0)^2 = 0/1(1) = 0/1 = 0

    why does my book mentions that answer is -2?
     
  2. jcsd
  3. Feb 7, 2009 #2
    huh, that was a stupid mistake, sorry.
    (I`m just totally nervous, I got an exam tomorrow)


    EDIT:
    but, how do I solve that anyway?

    I can`t manage to do it..

    I know the whole thing is about the "cos x" and that I have to make it turn to something in terms of y, but how?
     
  4. Feb 7, 2009 #3
    sec(x) is not 1/sin(x). It's 1/cos(x). Try the substitution u = cos(x). :smile:
     
  5. Feb 7, 2009 #4

    Tom Mattson

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    Staff Emeritus
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    Gold Member

    Not quite. If [itex]\sin^2(x)=y[/itex] then [itex]\cos^2(x)=1-y[/itex]. Here's the tricky part: [itex]\sec(x)=1/\sqrt{1-y}[/itex] if [itex]\cos(x)\geq0[/itex] but [itex]\sec(x)=-1/\sqrt{1-y}[/itex] if [itex]\cos(x)<0[/itex].
     
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