Results 1 to 4 of 4

Math Help - Calculating limits - general question

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    83

    Calculating limits - general question

    Hey!
    I encountered a question in my homework that was: "Evaluate the limit, if it exists".
    limit as x approaches 2 \frac{x^2 - x + 6}{x-2}
    It is not possible to factor out the denominator of this equation. I tried dividing (x-2) into the equation and got a solution of x+1R8.

    My question is: if I can't factor out the denominator of a rational equation, and the limit is asking where that equation would have the denominator at 0 (in the example 2) does this mean the limit does not exist?

    I hope I made sense in my question.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,820
    Thanks
    1711
    Awards
    1
    Yes, it means exactly that. The limit does not exist.

    BTW
    [tex] \displaystyle\lim _{x \to 2} \frac{{x^2 - 2x + 6}}{{x - 2}}[/tex] gives  \displaystyle\lim _{x \to 2} \frac{{x^2  - 2x + 6}}{{x - 2}} .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Kakariki View Post
    Hey!
    I encountered a question in my homework that was: "Evaluate the limit, if it exists".

    \displaystyle\lim_{x \to 2}\frac{x^2 - x + 6}{x-2}

    It is not possible to factor out the denominator of this equation. I tried dividing (x-2) into the equation and got a solution of x+1R8.

    My question is: if I can't factor out the denominator of a rational equation, and the limit is asking where that equation would have the denominator at 0 (in the example 2) does this mean the limit does not exist?

    I hope I made sense in my question.

    Thanks!
    You could simplify the fraction a little to

    \displaystyle\frac{x^2-x+6}{x-2}=\frac{x^2-x-2+8}{x-2}=\frac{(x-2)(x+1)+8}{x-2}

    =\displaystyle\frac{(x-2)(x+1)}{x-2}+\frac{8}{x-2}

    Hence the limit doesn't exist as the second fraction approaches \pm\infty as x approaches 2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,042
    Thanks
    1675
    Since you titled this "general question", here is a "general" answer.

    To find the limit \displaystyle\lim_{x\to a}\frac{P(x)}{Q(x)}, first try just putting x= a. If Q(a) is not 0, the limit is just \frac{P(a)}{Q(a)} (even if P(x)= 0). If Q(a)= 0 but P(a) is NOT 0, then the limit does not exist. If both P(a) and Q(a) are 0, then both P(a) and Q(a) must have a factor of x- a which can be cancelled.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: November 17th 2009, 10:27 PM
  2. Replies: 2
    Last Post: November 6th 2009, 11:45 AM
  3. Software for calculating general fractional factorial design?
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: March 31st 2009, 08:50 PM
  4. Calculating W for R-charts with prob. limits
    Posted in the Statistics Forum
    Replies: 0
    Last Post: October 16th 2008, 04:05 PM
  5. Calculating Limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 19th 2007, 05:32 PM

Search Tags


/mathhelpforum @mathhelpforum