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A question from limit

  1. Sep 29, 2005 #1
    :!!)

    lim(1-x).tan Pi X/2=???
    x->1

    please help meee :!!)
     
  2. jcsd
  3. Sep 29, 2005 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Let y=1-x and use tan=sin/cos and also sinu=cos(pi/2 -u), you will get your answer easily.
     
  4. Sep 30, 2005 #3
    thanks..
    one more guestion that I can't solve is...lim 1-sinX/cosx =?
    X->pi/2

    (I dont like trigonometric questions in limit)
    thanks:)
     
  5. Sep 30, 2005 #4

    VietDao29

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    Homework Helper

    [tex]\lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin x}{\cos x}[/tex]
    You can try multiplying both numerator and denominator by (1 + sin x), something like:
    [tex]= \lim_{x \rightarrow \frac{\pi}{2}} \frac{(1 - \sin x)(1 + \sin x)}{\cos x(1 + \sin x)}[/tex]
    [tex]= \lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin ^ 2 x}{\cos x(1 + \sin x)}[/tex]
    1 - sin2x = ...
    Can you go from here???
    ----------------------
    Or you can try a different way:
    [tex]1 - \sin x = \sin \left( \frac{\pi}{2} \right) - \sin x = 2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)[/tex]
    [tex]\lim_{x \rightarrow \frac{\pi}{2}} \frac{1 - \sin x}{\cos x}[/tex]
    [tex]= \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\sin\left( \frac{\pi}{2} - x \right)}[/tex]
    [tex]= \lim_{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{2\sin \left( \frac{\pi}{4} - \frac{x}{2} \right) \cos \left( \frac{\pi}{4} - \frac{x}{2} \right)}[/tex]
    [tex]= \lim_{x \rightarrow \frac{\pi}{2}} \frac{ \cos ^ 2 \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \cos \left( \frac{\pi}{4} - \frac{x}{2} \right)}[/tex]
    Viet Dao,
     
  6. Sep 30, 2005 #5
    thanks,yu are a good man :smile:
     
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