Results 1 to 3 of 3

Math Help - Evaluating limit

  1. March 21st 2010, 12:32 PM #1
    Neconine
    Neconine is offline
    Newbie
    Joined
    Nov 2009
    Posts
    11

    Evaluating limit

    Not sure how to go about this.

    Can I just rationalize either the numerator or denominator?

    lim (√2-x) - (√8-4x)
    x->2 (√3x+3) - (√11-x)
    Follow Math Help Forum on Facebook and Google+
  2. March 21st 2010, 01:37 PM #2
    running-gag
    running-gag is offline
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Quote Originally Posted by Neconine View Post
    Not sure how to go about this.

    Can I just rationalize either the numerator or denominator?

    lim (√2-x) - (√8-4x)
    x->2 (√3x+3) - (√11-x)
    Hi

    The numerator is \sqrt{2-x}-\sqrt{4(2-x)} = \sqrt{2-x}-2\sqrt{2-x} = -\sqrt{2-x}


    \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} = \frac{-\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}}\:\:\frac{\sqrt{3x+3}+\sqrt{11-x}}{\sqrt{3x+3}+\sqrt{11-x}}


    \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} = \frac{-\sqrt{2-x}\left(\sqrt{3x+3}+\sqrt{11-x}\right)}{4x-8}


    \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} = \frac{-\sqrt{2-x}\left(\sqrt{3x+3}+\sqrt{11-x}\right)}{-4(2-x)} = \frac{\sqrt{3x+3}+\sqrt{11-x}}{4\sqrt{2-x}}
    Follow Math Help Forum on Facebook and Google+
  3. March 21st 2010, 01:42 PM #3
    skeeter
    skeeter is offline
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,264
    Thanks
    1154
    Quote Originally Posted by Neconine View Post
    Not sure how to go about this.

    Can I just rationalize either the numerator or denominator?

    lim (√2-x) - (√8-4x)
    x->2 (√3x+3) - (√11-x)
    first, note that this limit should be one-sided since the domain of the rational expression is -1 \le x < 2

    \lim_{x \to 2^-} \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}}

    \lim_{x \to 2^-} \frac{\sqrt{2-x}-2\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}}

    \lim_{x \to 2^-} \frac{-\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}} \cdot \frac{\sqrt{3x+3}+\sqrt{11-x}}{\sqrt{3x+3}+\sqrt{11-x}}

    \lim_{x \to 2} \frac{-\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{ (3x+3)-(11-x)}

    \lim_{x \to 2^-} \frac{-\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{4x-8}

    \lim_{x \to 2} \frac{\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{8-4x}

    \lim_{x \to 2^-} \frac{\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{4(2-x)}

    \lim_{x \to 2^-} \frac{\sqrt{3x+3}+\sqrt{11-x}}{4\sqrt{2-x}}

    as x \to 2^- , the numerator approaches 6 and the denominator approaches 0 ... the limit does not exist.
    Follow Math Help Forum on Facebook and Google+
« application of definite integral | Problem with an integral »

Similar Math Help Forum Discussions

  1. Evaluating limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 3rd 2010, 07:03 AM
  2. Evaluating a limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 2nd 2010, 02:01 PM
  3. evaluating a limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 12th 2009, 06:01 PM
  4. evaluating a limit
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: October 17th 2009, 10:54 PM
  5. Evaluating A Limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 1st 2009, 11:43 PM

Search Tags


/mathhelpforum @mathhelpforum