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Lim x->0 (arctan(x^2))/(sinx+tanx)^2

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data

    lim x→0 [itex]\frac{arctan(x^{2})}{(sinx+tanx)^{2}}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    OK, so at first, the form is [itex]\frac{0}{0}[/itex] so I use L'Hopital's rule. I get the following as my answer.

    lim x→0 [itex]\frac{1}{1+x^{4}}[/itex]*[itex]\frac{1}{2(sinx+tanx)(cosx+sec^{2}x}[/itex]

    But the form is not 0/0 and I cannot figure out what to do at this point. I have tried expanding out the (sinx+tanx) with the (cosx+sec^2x) to no avail.. and I can't really think of what else to do.

    Thanks in advance!
     
  2. jcsd
  3. Dec 1, 2011 #2
  4. Dec 1, 2011 #3

    Dick

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    You did not differentiate arctan(x^2) correctly. Don't forget the chain rule!
     
  5. Dec 1, 2011 #4
    Thanks! But now I have another issue.

    When I fix my error, I get

    lim x→0 [itex]\frac{x}{(1+x^4)(sinx+tanx)(cosx+sec^2x)}[/itex]

    This is of the form 0/0

    I tried to expand the bottom and I get a HUGE equation, a very very long equation and I must be doing this wrong. Any ideas?

    PS: This is the equation I get.

    [itex]\frac{x}{sinxcosx+sinxsec^{2}x+sinx+tanxsec^{2}x+x^{4}sinxcosx+x^{4}sinxsec^{2}x+x^{4}sinx+x^{4}tanxsec^{2}x}[/itex]

    I really doubt that I am expected to use hopitals rule on that.. even though it wouldn't be too hard
     
  6. Dec 1, 2011 #5

    Dick

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    What's lim x->0 of 1/((1+x^4)*(cos(x)+sec(x)^2))? That part is not indeterminant. It just approaches a constant. Just do l'Hopital on the indeterminant part.
     
  7. Dec 1, 2011 #6
    Oh, I completely forgot you could do that! I get 1/4 is the answer to the question.. which I hope is right. And thanks so much for your help!
     
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