Page 1 of 2 12 LastLast
Results 1 to 15 of 20

Math Help - limit n-->infinty

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    49

    limit n-->infinty

    Hi my friends
    can you help to find the limit of the following equation

    Area = \lim_{n \rightarrow \infty} \frac{\pi}{n} *sin(\frac{\pi*n+\pi}{2})
    Last edited by Jameson; December 23rd 2009 at 03:44 PM. Reason: fixed Latex
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Well think about it. \left|\sin\left(\frac{\pi(n+1)}{2}\right)\right| is always less than or equal to 1, and \frac{\pi}{n} gets as small as you wish...
    Last edited by Bruno J.; December 21st 2009 at 07:45 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2009
    Posts
    69
    Quote Originally Posted by metallica007 View Post
    Hi my friends
    can you help to find the limit of the following equation

    Area = \lim n--> \infty \frac{\pi}{n} *sin(\frac{\pi*n+\pi}{2})
    Well u can use the squeeze theorem.

    -1<= sinx<=1
    so
    lim n => inf -pi/n<= pi/n *sinx<=pi/n
    From this, is pretty obvious what the answer is...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2009
    Posts
    49
    Guys your killing me
    I did not get it
    i know this :
    lim n--> infinity pi/n =0
    lim n--> infinity sin(\frac{\pi*n+\pi}{2}) does not exist
    so the answer is ... 0 !!!

    please correct me if i am wrong
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2009
    Posts
    69
    Quote Originally Posted by metallica007 View Post
    Guys your killing me
    I did not get it
    i know this :
    lim n--> infinity pi/n =0
    lim n--> infinity sin(\frac{\pi*n+\pi}{2}) does not exist
    so the answer is ... 0 !!!

    please correct me if i am wrong
    Sin always exist between -1 and 1...
    and answer is correct
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2009
    Posts
    49
    but the answer doesn't seems right to me
    because i tried to find the area of sin(x) from 0 to pi using limit --> infinity and Riemann sums and i got 0
    but when i tried to integrate it using integration rules and i get 2 !!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2009
    Posts
    49
    i am , doing it like this
    \Delta x = \frac{\pi}{n}

    \lim n-->\infty \sum Sin(xi) \Delta x

    \lim n-->\infty\\\\   Sin(\frac{\pi}{n} \frac{n(n+1)}{2}) \frac{\pi}{n}

    \lim n--> \infty \frac{\pi}{n} *sin(\frac{\pi n+\pi}{2})

    I dont know if that was the right steps to find area using Riemann sum and limit

    please correct me if i am wrong
    Follow Math Help Forum on Facebook and Google+

  8. #8
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    0... The answer is 0...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Dec 2009
    Posts
    49
    Quote Originally Posted by VonNemo19 View Post
    0... The answer is 0...
    when a =0 , b = pi

    does \int\sin(x)= 0 !!!!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by metallica007 View Post
    when a =0 , b = pi

    does \int\sin(x)= 0 !!!!
    No, actually its 1
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by VonNemo19 View Post
    No, actually its 1
    2
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Dec 2009
    Posts
    49
    Guys , what about my question
    why i get 0 instead of 2 ( or 1 ) !!!!
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,700
    Thanks
    454
    Quote Originally Posted by metallica007 View Post
    Guys , what about my question
    why i get 0 instead of 2 ( or 1 ) !!!!
    as stated before ...

    -1 \le \sin(anything) \le 1

    so ...

    \lim_{n \to \infty} \, \frac{\pi}{n}(a \, value \, stuck \, between \, -1 \, and \, 1) = 0 <br />

    now look at the graph
    Attached Thumbnails Attached Thumbnails limit n--&gt;infinty-singraf.jpg  
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,673
    Thanks
    1618
    Awards
    1
    Quote Originally Posted by metallica007 View Post
    can you help to find the limit of the following equation
    Area = \lim _{n \to \infty }\frac{\pi}{n} *sin(\frac{\pi*n+\pi}{2})
    It seems to me that this thread is very confused.
    It seem that the OP assumes that the above limit is an approximating sum for \int\limits_0^\pi  {\sin (x)dx} .
    But that is totally wrong. A correct approximating sum (righthand sum) is \frac{\pi }{n}\sum\limits_{k = 1}^n {\sin \left( {\frac{\pi }{n}k} \right)} .
    Thus we want this limit: \lim _{n \to \infty } \frac{\pi }{n}\sum\limits_{k =1 }^n {\sin \left( {\frac{\pi }{n}k} \right)} .
    Last edited by Plato; December 23rd 2009 at 04:44 PM. Reason: minor typo
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,700
    Thanks
    454
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Limits involving products of infinty and zero.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 4th 2011, 12:32 PM
  2. Finding the limit as n goes to infinty
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 26th 2010, 12:41 AM
  3. product of n unifom distribution as n-> infinty
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: November 11th 2009, 10:35 AM
  4. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM
  5. continuous at infinty?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 11th 2008, 09:10 PM

Search Tags


/mathhelpforum @mathhelpforum