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Thread: Trig Limits

  1. #1
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    Trig Limits

    I'm trying to evaluate this limit and i'm having some issues.

    $\displaystyle \lim_{x\to 0} \frac{1-\sec^2(2x)}{x^2} $

    I'm not 100% sure about my trig Identities but i think that
    $\displaystyle 1-\sec^2(2x) $ is equal to $\displaystyle 2\tan^2(x) $
    is that correct?

    if so, i say that the equation is equal to $\displaystyle \frac{2\sin^2(x)}{2\cos^2(x)}(\frac{1}{x^2}) $

    I have a feeling this is wrong somehow knowing the answer is -4. Can someone help me?

    thank you
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  2. #2
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    Quote Originally Posted by discobob View Post
    I'm trying to evaluate this limit and i'm having some issues.

    $\displaystyle \lim_{x\to 0} \frac{1-\sec^2(2x)}{x^2} $

    I'm not 100% sure about my trig Identities but i think that
    $\displaystyle 1-\sec^2(2x) $ is equal to $\displaystyle 2\tan^2(x) $
    is that correct?

    if so, i say that the equation is equal to $\displaystyle \frac{2\sin^2(x)}{2\cos^2(x)}(\frac{1}{x^2}) $

    I have a feeling this is wrong somehow knowing the answer is -4. Can someone help me?

    thank you
    The identity is $\displaystyle sec^2(\theta)=1+tan^2(\theta)$.
    for any well-defined $\displaystyle \theta$.
    Now, Try to correct your mistake.

    Hint: Try to get the minus sign as a common factor from the numerator.
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  3. #3
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    ok, this sounds so easy but i'm not so sure on how to get . That should be pretty basic but nothing is coming to mind. I mean, i know you can factor out a -1 getting $\displaystyle -1(-1+\sec^2(2x)) $. But that's not what you mean right?
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  4. #4
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    $\displaystyle

    lim_{x->0} ((1-sec˛2x)/(x˛))
    = lim_{x->0}((cos˛2x-1)/(x˛))
    = lim_{x->0}((-(1-cos˛2x)4)/((2x)˛))
    =lim_{x->0}(((sin˛2x)(-4))/((2x)˛))
    =-4


    $
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  5. #5
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    $\displaystyle

    lim_{x->0} ((1-sec˛2x)/(x˛))
    = lim_{x->0}((cos˛2x-1)/(x˛))
    = lim_{x->0}((-(1-cos˛2x)4)/((2x)˛))
    =lim_{x->0}(((sin˛2x)(-4))/((2x)˛))
    =-4

    $
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  6. #6
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    lim_{x->0} ((1-sec˛2x)/(x˛))
    = lim_{x->0}((cos˛2x-1)/(x˛))
    = lim_{x->0}((-(1-cos˛2x)4)/((2x)˛))
    =lim_{x->0}(((sin˛2x)(-4))/((2x)˛))
    =-4
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  7. #7
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    i forgot the cos^2(2x) in the denominator.nevertheless that would vanish as cos 0=1 after putting the limits.
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