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Math Help - limits tan^2x and cot^x

  1. #1
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    limits tan^2x and cot^x



    I split it up into the limit as x --> pi of

    tan^2x + cot^2x

    tan^2x = 1 - 1/cos^2x

    in the limit x --> pi 1--1 = 2

    for the cot^2 one

    cot^2x = 1/tan^2x

    i used the double angle formula and simplified to

    \frac{1}{1-\frac{2tanx}{tan2x}}

    then the infinities of tanx/tan2x cancel out so your left with 1/(1-2) = -1

    i don't know how to more formally do the cot one!
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  2. #2
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    Re: limits tan^2x and cot^x

    [QUOTE=Applestrudle;817376]

    I split it up into the limit as x --> pi of

    tan^2x + cot^2x[quote]
    ??? \frac{cos^2(x)}{sin^2(x)}= cot^2(x), not " tan^2(x)" and \frac{cos^3(x)}{sin^2(x)}= cos(x)cot^2(x), not cot^2(x). Did you not notice that, at x=\frac{\pi}{4} the first is 1+ \frac{\sqrt{2}}{2} but you "split" formula has value 2?
    I think you would be better advised to write it as cos^2(x)\left(\frac{1+ cos(x)}{sin^2(x)}\right)= cos^2(x)\left(\frac{1+ cos(x)}{1- cos^2(x)}\right).

    tan^2x = 1 - 1/cos^2x

    in the limit x --> pi 1--1 = 2

    for the cot^2 one

    cot^2x = 1/tan^2x

    i used the double angle formula and simplified to

    \frac{1}{1-\frac{2tanx}{tan2x}}

    then the infinities of tanx/tan2x cancel out so your left with 1/(1-2) = -1

    i don't know how to more formally do the cot one!
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  3. #3
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    Re: limits tan^2x and cot^x

    [QUOTE=HallsofIvy;817379][QUOTE=Applestrudle;817376]

    I split it up into the limit as x --> pi of

    tan^2x + cot^2x
    ??? \frac{cos^2(x)}{sin^2(x)}= cot^2(x), not " tan^2(x)" and \frac{cos^3(x)}{sin^2(x)}= cos(x)cot^2(x), not cot^2(x). Did you not notice that, at x=\frac{\pi}{4} the first is 1+ \frac{\sqrt{2}}{2} but you "split" formula has value 2?
    I think you would be better advised to write it as cos^2(x)\left(\frac{1+ cos(x)}{sin^2(x)}\right)= cos^2(x)\left(\frac{1+ cos(x)}{1- cos^2(x)}\right).
    okay a lot of stupid mistakes,

    I get cot^2x goes to zero and cosxcot^2x goes to zero too
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