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Limits of Functions and Asymptotes

  1. Feb 3, 2010 #1
    1. I am concerned with finding the discontinuities of a functin say x3+3x2+2x / (x-x3)



    2. I am having issues with classifying for the type of discontinuities. Finding them is not an issue.



    3. Also when finding horizontal asymptotes F(x) = 4 / (2e-x +1) I understnad that the HA are found by taking the limit of the function as X--->-INF/INF , but why is one of the HA 4?

    Thank You =).
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 3, 2010 #2

    Mark44

    Staff: Mentor

    Do you mean classifying them into vertical or horizontal asymptotes? The vertical asymptotes are generally the numbers that make the denominator zero, that don't also make the numerator zero. To find them, factor both the numerator and denominator. The numbers that make the denominator zero are x = -1, x = 0, and x = 1. x = 0 is NOT a vertical asymptote, because the numerator is also zero when x = 0.
    As x --> infinity, e-x --> 0, so the denominator --> 1, and the overall fraction --> 4.
     
  4. Feb 3, 2010 #3
    Thank You for your help. I was actually meaning the discontinuities as Jump,Removable, Infinite. So far I have understood that only piece wise functions can have a Jump. Infinite limits are when the lim of F(x) as X-->A = INF/-INF. And removable is when the limit exists , but is not defined. Am I correct?
     
  5. Feb 3, 2010 #4

    Mark44

    Staff: Mentor

    Close. A removable discontinuity occurs when [tex]lim_{x \to a} f(x)[/tex] exists (that means both one-sided limits exist), but f is not defined at a.

    In your first example, I believe that there is a removable discontinuity at x = 0.
     
  6. Feb 3, 2010 #5
    Yea, that is correct. =) Thank you sir.
     
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