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Math Help - Limits

  1. #1
    Member roshanhero's Avatar
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    Limits

    lim.....asinx-sin2x/tan^3x is finite,find the value of 'a' and find the limit.
    x->0
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    We can rewrite as:

    \lim_{x\to 0}\frac{(a-2cos(x))cos^{3}(x)}{sin^{2}(x)}

    Look at a-2cos(x)

    What if a=2, what is the limit?.
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  3. #3
    MHF Contributor
    skeeter's Avatar
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    I'm assuming that this is your limit expression ...

    \lim_{x \to 0} \frac{a\sin{x} - \sin(2x)}{\tan^3{x}}

    using L'Hopital ...

    \lim_{x \to 0} \frac{a\cos{x} - 2\cos(2x)}{3\tan^2{x}\sec^2{x}}

    at this point, substituting 0 in the denominator yields 0 and substituting 0 in the numerator yields (a - 2). for the limit to exist, a must equal 2.

    starting from scratch with a = 2 ...

    \lim_{x \to 0} \frac{2\sin{x} - \sin(2x)}{\tan^3{x}}

    \lim_{x \to 0} \frac{2\sin{x} - 2\sin{x}\cos{x}}{\tan{x}(\sec^2{x} - 1)}

    \lim_{x \to 0} \frac{2\sin{x}(1 - \cos{x})}{\tan{x}(\sec^2{x} - 1)}

    \lim_{x \to 0} \frac{2\sin{x}(1 - \cos{x})}{\tan{x}(\sec{x}+1)(\sec{x}-1)}

    \lim_{x \to 0} \frac{2\sin{x}(1 - \cos{x})}{\frac{\sin{x}}{\cos{x}}(\sec{x}+1)\left(  \frac{1-\cos{x}}{\cos{x}}\right)}

    \lim_{x \to 0} \frac{2}{\sec^2{x}(\sec{x}+1)} = 1
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