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Math Help - Help with L'Hopital's Rule?

  1. #1
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    Help with L'Hopital's Rule?

    I need to evaluate these limits using L'Hopital's Rule, but i do not understand how to set these up. Full answers would be extremely helpful, but i could deal with set ups.
    1.
    lim
    n->infinity

    n(1-cos(1/n))

    and
    2.
    lim
    x->(pi/2) (left)

    (x-(pi/2))sec(x)
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  2. #2
    Super Member

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    Hello, jagger04!

    1)\;\;\lim_{x\to\infty}n\left(1-\cos\tfrac{1}{n}\right)

    We have: . \lim_{x\to\infty}\frac{1-\cos\frac{1}{n}}{\frac{1}{n}} \;\to\; \frac{0}{0}

    Apply l'Hopital: . \lim_{x\to\infty}\frac{\left(\text{-}\sin\frac{1}{n}\right)\left(\text{-}\frac{1}{n^2}\right)}{\text{-}\frac{1}{n^2}} \;=\;\lim_{x\to\infty}\left(\text{-}\sin\frac{1}{n}\right) \;=\;\text{-}\sin0 \;=\;0




    2)\;\;\lim_{x\to\frac{\pi}{2}} \left(x -\tfrac{\pi}{2}\right)\sec x

    We have: . \lim_{x\to\frac{\pi}{2}}\frac{\left(x - \frac{\pi}{2}\right)}{\cos x} \;\to \;\frac{0}{0}

    Apply L'Hopital: . \lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\text{-}\sin x}\right) \;=\;\frac{1}{\text{-}1} \;=\;-1

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