Results 1 to 4 of 4

Math Help - Need help with limit

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    1

    Need help with limit

    Hello,

    I am having trouble finding the following limit:

    lim as x -> infinity of sin(x-1/x^2+2)

    Factoring the denominator does not help.

    Please help?

    Thanks so much
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by tigerlily View Post
    Hello,

    I am having trouble finding the following limit:

    lim as x -> infinity of sin(x-1/x^2+2)

    Factoring the denominator does not help.

    Please help?

    Thanks so much
    \frac{x-1}{x^2 - 1} = \frac{1}{x+1} \to 0 as x \to +\infty. And you should have a theorem about the limit of a composite functions that you can use ....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,785
    Thanks
    1569
    Quote Originally Posted by mr fantastic View Post
    \frac{x-1}{x^2 - 1} = \frac{1}{x+1} \to 0 as x \to +\infty. And you should have a theorem about the limit of a composite functions that you can use ....
    I don't see how this helps, considering \displaystyle \frac{x-1}{x^2 - 1} \neq \frac{x - 1}{x^2 + 2}...


    Instead, \displaystyle \lim_{x \to \infty}\frac{x - 1}{x^2 + 2} = \lim_{x \to \infty}\frac{\frac{1}{x} - \frac{1}{x^2}}{1 + \frac{2}{x^2}} = 0.

    And from the continuity of the sine function

    \displaystyle \lim_{x \to \infty}\sin{\left(\frac{x-1}{x^2 + 2}\right)} = \sin{\left[\lim_{x \to \infty}\left(\frac{x-1}{x^2 + 2}\right)\right]} = \sin{0} = 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Prove It View Post
    I don't see how this helps, considering \displaystyle \frac{x-1}{x^2 - 1} \neq \frac{x - 1}{x^2 + 2}...


    Instead, \displaystyle \lim_{x \to \infty}\frac{x - 1}{x^2 + 2} = \lim_{x \to \infty}\frac{\frac{1}{x} - \frac{1}{x^2}}{1 + \frac{2}{x^2}} = 0.

    And from the continuity of the sine function

    \displaystyle \lim_{x \to \infty}\sin{\left(\frac{x-1}{x^2 + 2}\right)} = \sin{\left[\lim_{x \to \infty}\left(\frac{x-1}{x^2 + 2}\right)\right]} = \sin{0} = 0
    I could have sworn it was x^2 - 1 just before I posted.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 8th 2010, 12:29 PM
  2. Replies: 1
    Last Post: February 5th 2010, 04:33 AM
  3. Replies: 16
    Last Post: November 15th 2009, 05:18 PM
  4. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 06:05 PM
  5. Replies: 15
    Last Post: November 4th 2007, 08:21 PM

Search Tags


/mathhelpforum @mathhelpforum