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Help with limit

  1. Sep 15, 2008 #1
    [tex]x\stackrel{lim}{\rightarrow}pi[/tex] sin(x) / (pi - x)

    Sorry about the awkward looking notation, trying to understand how to use this stuff ^^

    Anyways, any idea on how I can solve this? I am totally stumped. I know that sin(x) / x is one, but that damn pi symbol totally throws me :P

    Thanks ^^
    Adrian
     
  2. jcsd
  3. Sep 15, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi Adrian! Welcome to PF! :smile:

    (have a pi: π :smile:)

    Two methods:

    i] since you know the limit for sin(x)/x, just put y = π - x :wink:

    ii] use the same method that you would have used to find sin(x)/x :smile:

    (oh … and to get [tex]x\stackrel{lim}{\rightarrow}\pi \frac{sin x}{\pi - x}[/tex],

    type [noparse][tex]x\stackrel{lim}{\rightarrow}\pi \frac{sin x}{\pi - x}[/tex][/noparse])
     
  4. Sep 15, 2008 #3
    Well that's actually indeterminant, isn't it?
     
  5. Sep 15, 2008 #4

    HallsofIvy

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    Since sin(x)/x is not "indeterminant" why would sin(x)/(pi- x) be? Just putting x= pi gives the "indeterminant" 0/0 which just means we can't do that to find the limit. There is nothing "indeterminant" about the limit.

    adoado, you can also use sin(pi- x)= - sin(x) to rewrite sin(x)/(pi- x) as -sin(pi- x)/(pi-x). NOW replace "pi- x" by "y".
     
  6. Sep 15, 2008 #5
    You don't need "\stackrel":


    [tex]\lim_{x\rightarrow\pi} \frac{\sin(x)}{\pi - x}=1[/tex]
     
  7. Sep 15, 2008 #6
    I meant that since it's in an indeterminant form, 0/0, l'hopital's rule comes into play.
     
  8. Sep 15, 2008 #7

    HallsofIvy

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    That depends upon what you mean by "comes into play". In this case, the problem is much too simple to require L'Hopital's rule and I suspect that the OP has not yet had L'Hopital's rule.
     
  9. Sep 15, 2008 #8
    I mean you can use L'Hopital's rule to solve this. Either way is straight forward, but sometimes it's nice to know more than one way to solve a problem.
     
  10. Sep 16, 2008 #9
    Hello ^^

    Thanks everyone, I think I understand it now. And thanks for the syntax corrections :P

    And about L'Hopitals Rule, indeed I have not learnt it but its a good thing to know ;)

    Thanks,
    Adrian
     
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