Results 1 to 4 of 4

Math Help - special limits of trig functions

  1. #1
    Junior Member
    Joined
    Nov 2006
    Posts
    41

    special limits of trig functions

    compute using special limits of trig functions: lim x->pie (cosx+1)/(x-pie)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,815
    Thanks
    315
    Awards
    1
    Quote Originally Posted by myoplex11 View Post
    compute using special limits of trig functions: lim x->pie (cosx+1)/(x-pie)
    First of all, it's "pi" not "pie."

    I'm not sure how to break this into anything resembling a "special limit" but it's not hard to do directly:

    \lim_{x \to \pi} \frac{cosx + 1}{x - \pi}

    Note that this is of the form 0/0, so use L'Hopital's rule:

    \lim_{x \to \pi} \frac{cosx + 1}{x - \pi} = \lim_{x \to \pi} \frac{-sinx}{1} = 0

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by topsquark View Post
    First of all, it's "pi" not "pie."

    I'm not sure how to break this into anything resembling a "special limit" but it's not hard to do directly:

    \lim_{x \to \pi} \frac{cosx + 1}{x - \pi}

    Note that this is of the form 0/0, so use L'Hopital's rule:

    \lim_{x \to \pi} \frac{cosx + 1}{x - \pi} = \lim_{x \to \pi} \frac{-sinx}{1} = 0

    -Dan
    You can do this,
    \lim_{x\to \pi} \frac{1-\cos (x- \pi)}{x-\pi}
    Using the identity,
    \cos x=-\cos (x-\pi)
    Now this is a composition function,
    f(x)=\frac{1-\cos x}{x}
    g(x)=x-\pi
    And,
    \lim_{x\to \pi}g(x)=0
    Thus the limit of,
    \lim_{x\to \pi}\frac{1-\cos (x-\pi)}{x-\pi}=\lim_{x\to 0}\frac{1-\cos x}{x}=0
    By composition rule for limits.
    That is a standard limit.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,815
    Thanks
    315
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    You can do this,
    \lim_{x\to \pi} \frac{1-\cos (x- \pi)}{x-\pi}
    Using the identity,
    \cos x=-\cos (x-\pi)
    Now this is a composition function,
    f(x)=\frac{1-\cos x}{x}
    g(x)=x-\pi
    And,
    \lim_{x\to \pi}g(x)=0
    Thus the limit of,
    \lim_{x\to \pi}\frac{1-\cos (x-\pi)}{x-\pi}=\lim_{x\to 0}\frac{1-\cos x}{x}=0
    By composition rule for limits.
    That is a standard limit.
    Clever! I like it.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limits of Trig Functions
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 5th 2010, 11:58 AM
  2. Limits to Trig Functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 18th 2010, 01:24 AM
  3. help with special trig limits
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: October 25th 2008, 12:18 PM
  4. Limits using Trig Functions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 22nd 2008, 01:12 PM
  5. Limits of Trig Functions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 12th 2007, 07:22 PM

Search Tags


/mathhelpforum @mathhelpforum