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Difficulty to find this limit of trigonometric function

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello! I'm having some trouble to find this limit [tex]\lim_{x\rightarrow \pi}\frac{senx}{x-\pi }[/tex] Also, limits like this one [tex]\lim_{x\rightarrow \frac{\pi}2{}}\frac{1-senx}{2x-\pi }[/tex] confuse me a lot!


    2. Relevant equations



    3. The attempt at a solution
    I don't even know where to start! I tried this but it's problably just waste of time.
    My idea was try calculate the area of the third sector of the trigonometric circle. Because to me that -∏ is making my area goes to the third sector, the negative one. Using this idea I've done:
    [tex]\frac{-senx}{senx}< \frac{-x}{senx}<\frac{-tgx}{senx}[/tex]
    Dividing by senx we got
    [tex]-1> \frac{senx}{x}>-1[/tex]
    But I'm stuck here. Help please! :wink:

    And forgive me for my bad english! English is not my native language.
     
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 23, 2012 #2

    Mark44

    Staff: Mentor

    You can rewrite this as
    $$\lim_{x \to \pi}\frac{sin(\pi - x)}{x - \pi}$$

    Then, let u = x - ##\pi##.

    There's a very common limit that will come in handy here.
     
  4. Apr 23, 2012 #3

    Mark44

    Staff: Mentor

    BTW, in English we call it sine (or sin).
     
  5. Apr 23, 2012 #4
    Sorry, my bad!

    Thanks for your answer but I still don't get it! Why can I rewrite my limit as?
    [tex]\lim_{x\rightarrow \pi } \frac{sin(\pi -x)}{x-\pi}[/tex]
    Well, assuming that it's ok to rewrite the limit as above I think I know what to do next:
    [tex]\lim_{x\rightarrow \pi } \frac{sin(-(\pi -x))}{x-\pi}[/tex]
    [tex]\lim_{x\rightarrow \pi } \frac{-sin(x-\pi)}{x-\pi}[/tex]
    [tex]\lim_{x\rightarrow \pi } \frac{-sin(x-\pi)}{x-\pi}=-1[/tex]
     
  6. Apr 23, 2012 #5

    Mark44

    Staff: Mentor

    1. Because sin(x) = sin(##\pi## - x)
    2. If u = x - ##\pi##, then as x approaches ##\pi##, u approaches 0. Your limit is ##\lim_{u \to 0} \frac{-sin(u)}{u}. ##
     
  7. Apr 23, 2012 #6

    Mark44

    Staff: Mentor

    For the other problem, a similar approach will work, I believe.
     
  8. Apr 23, 2012 #7
    I just didn't get why sin(x) = sin(π - x). I tried to solve this one by doing what you did [tex]\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-sinx}{2x-\pi }[/tex]
    this is result
    [tex]\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-sinx}{2x-\pi }=\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-sin(\pi -x)}{2x-\pi }=\lim_{x\rightarrow \frac{\pi }{2}}\frac{1+sin(x-\pi )}{2x-\pi }\frac{2}{2}=\lim_{x\rightarrow \frac{\pi }{2}}\frac{2+2sin(x-\pi)}{4x-2\pi}=\lim_{x\rightarrow \frac{\pi }{2}}\frac{2+2sin(x-\pi)}{2(2x-\pi)}=\lim_{x\rightarrow \frac{\pi }{2}}\frac{2}{2}-\frac{1}{1} = 0[/tex]
    This is totally wrong :grumpy:
     
  9. Apr 23, 2012 #8

    Mark44

    Staff: Mentor

    sin(x) = cos(##\pi##/2 - x)

    2x - ##\pi## = 2(x - ##\pi##/2)
     
  10. Apr 23, 2012 #9
    I still don't know how to solve [tex]\lim_{x\rightarrow \frac{\pi}2{}}\frac{1-senx}{2x-\pi }[/tex]
     
  11. Apr 23, 2012 #10

    Mark44

    Staff: Mentor

    Post #8 has some clues. Do you know of any special trig limits?
     
  12. Apr 23, 2012 #11
    I know about senx/x.
    Using your clues:
    [tex]\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-cos(x-\frac{\pi }{2})}{2(x-\frac{\pi }{2})} = \lim_{x\rightarrow \frac{\pi }{2}} \frac{1-1}{2}=0[/tex]
     
  13. Apr 23, 2012 #12

    Mark44

    Staff: Mentor

    You have a mistake. The denominator is approaching 0, not 2.

    The other limit I referred to is
    $$\lim_{t \to 0}\frac{1 - cos(t)}{t} = 0$$
     
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