1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A question from limit

  1. Sep 29, 2005 #1

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

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


    User Avatar
    Science Advisor

    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
    one more guestion that I can't solve is...lim 1-sinX/cosx =?

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


    User Avatar
    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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook