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Math Help - Which trig limit did they use?

  1. #1
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    Which trig identity did they use?

    The book I'm using seriously skips soooo many steps - it's hard to figure out how they get from one step to another.
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  2. #2
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    Hello, RedSwiss!

    The book I'm using seriously skips soooo many steps.
    It's hard to figure out how they get from one step to another.

    . . \displaystyle \lim_{x\to\infty}\frac{1-\cos\left(\frac{1}{x}\right)}{\frac{1}{x^2}} \;=\;\lim_{x\to\infty}\frac{-\frac{1}{x^2}\sin\left(\frac{1}{x}\right)}{-\frac{2}{x^3}}

    They used L'Hopital's Rule.

    We have: .  \dfrac{1-\cos(x^{-1})}{x^{-2}}


    The derivative of the numerator is: . -\bigg[-\sin\left(x^{-1}\right)\bigg]\bigg[-x^{-2}\bigg] \;=\;-\frac{1}{x^2}\sin\left(\frac{1}{x}\right)

    The derivative of the denominator is: . -2x^{-3} \;=\;-\frac{2}{x^3}

    Last edited by Soroban; August 27th 2010 at 10:48 PM.
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  3. #3
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    in this problem L'Hospital's rule has been used. that is how you get the second expression from the first. according to LHospital's rule lim(x->a) n(x)/d(x)=lim(x->a) n'(x)/d'(x)=............=lim(x->a) n''''''....(x)/d'''''.....(x). differentiation can be carried out untill the undefined form of the given expression is got rid of.
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