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Math Help - lhospital (finals my brain is mush)

  1. #1
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    lhospital (finals my brain is mush)

    Why can't I see this I don't know...

    lim( as x approaches pi/2 from the left) (secx - tanx)

    can't use math tags worth a damn.

    Im pretty sure this is lhospital's rule because just taking the limit gives me 1 + infinite which is not the correct answer, I converted with the natural log and did common denominator but I cant think anymore.. studying for 5 finals (calc and physics being 2) just makes my brain melt
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  2. #2
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    \lim_{x \to \frac{\pi}{2}^+} \left( \sec x - \tan x\right) \ = \ \lim_{x \to \frac{\pi}{2}^+}\left( \frac{1}{\cos x} - \frac{\sin x}{\cos x}\right) \ = \ \lim_{x \to \frac{\pi}{2}^+} \frac{1 - \sin x}{\cos x} = \left[\frac{0}{0}\right]

    Now use L'Hopital's.
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  3. #3
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    Quote Originally Posted by ryukolink View Post
    Why can't I see this I don't know...

    lim( as x approaches pi/2 from the left) (secx - tanx)

    can't use math tags worth a damn.

    Im pretty sure this is lhospital's rule because just taking the limit gives me 1 + infinite which is not the correct answer, I converted with the natural log and did common denominator but I cant think anymore.. studying for 5 finals (calc and physics being 2) just makes my brain melt
    Hi ryukolink,

    lim_{x \rightarrow \frac{\pi}{2}^-}(\sec x - \tan x) =lim_{x \rightarrow \frac{\pi}{2}^-}(\frac{1}{\cos x}-\frac{\sin x}{\cos x}) =lim_{x \rightarrow \frac{\pi}{2}^-}(\frac{1-\sin x}{\cos x}) =lim_{x \rightarrow \frac{\pi}{2}^-}(\frac{-\cos x}{-\sin x}) (Since you got zero over zero)
    =lim_{x \rightarrow \frac{\pi}{2}^-}(\cot x)=?
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  4. #4
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    Oh my gosh, sec is 1/cos what was I thinking, and I am almost in linear and differential, holy moley.. I am a gumby thank you.
    Last edited by mr fantastic; March 2nd 2009 at 03:40 AM. Reason: Modified language
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