Results 1 to 7 of 7

Math Help - One sided Limits

  1. #1
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32

    One sided Limits

    I was posed this problem on another forum by a first year and for some reason I'm struggling

    lim x-> 1 from the left of
    (1-2x)/(x^2-1)

    Could someone provide me a precise way to show the limit goes to infinity?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Oct 2008
    Posts
    24
    I use a theorem : let {x_k}->1 from left as k->infinite , and f be continuous if {f(x_k)}->infinite , limf(x)=infiniteas x->1^-

    clearly , (1-2x)/(x^2-1) is a continuous on (-infinite , 1^-) and let {x_k}={(k-1)/k}converges to 1 from left.

    And f(x_k)={(k^2)/2}-k/(4k-2) this is infinite as k->infinite
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32
    Quote Originally Posted by chipai View Post
    I use a theorem : let {x_k}->1 from left as k->infinite , and f be continuous if {f(x_k)}->infinite , limf(x)=infiniteas x->1^-

    clearly , (1-2x)/(x^2-1) is a continuous on (-infinite , 1^-) and let {x_k}={(k-1)/k}converges to 1 from left.

    And f(x_k)={(k^2)/2}-k/(4k-2) this is infinite as k->infinite
    Oh yes! I have heard a theorem like that before. So I follow you up until the point where you let {x_k}={(k-1)/k}. It is an increasing function, but is there a way to find such {x_k}? EDIT: or is it arbitrary?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    24
    Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32
    Quote Originally Posted by chipai View Post
    Since a divergent sequence can has a convergent subsequence ,the sequence i used is not arbitary.
    thank you for your help.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    think simple: x^2 - 1 approaches 0 from negative values because x \rightarrow 1-, and 1 - 2x approaches -1. so the limit is +\infty.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member MacstersUndead's Avatar
    Joined
    Jan 2009
    Posts
    291
    Thanks
    32
    Quote Originally Posted by NonCommAlg View Post
    think simple: x^2 - 1 approaches 0 from negative values because x \rightarrow 1-, and 1 - 2x approaches -1. so the limit is +\infty.
    Well, I was afraid that such an answer was much too simple. :P
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. One sided limits?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 14th 2011, 10:11 PM
  2. One Sided Limits
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 14th 2009, 04:18 AM
  3. One Sided Limits
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 9th 2009, 09:23 PM
  4. one-sided limits
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 8th 2009, 09:23 PM
  5. need help plz sided limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 7th 2006, 08:58 AM

Search Tags


/mathhelpforum @mathhelpforum