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Divergent limit + divergent limit = convergent limit . Is it possible?

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data

    As a part of Method of Frobenius, I am encountered with the following problems:

    Evaluate the following limits:

    Q1. [itex]\stackrel{limit}{_{x→0}}\frac{1-2x}{x}[/itex]

    Q2. [itex]\stackrel{limit}{_{x→0}}\frac{x-1}{x}[/itex]

    Q3. [itex]\stackrel{limit}{_{x→0}}\frac{1-2x}{x}+\frac{x-1}{x}[/itex]

    In context of the above problems, I am having difficulty in verifying the following property of limits:

    [itex]\stackrel{limit}{_{x→a}}(f(x)+g(x))=\stackrel{limit}{_{x→a}}f(x)+ \stackrel{limit}{_{x→a}}g(x)[/itex]​

    2. Relevant equations

    3. The attempt at a solution

    A1. [itex]\stackrel{limit}{_{x→0}}\frac{1-2x}{x}=∞[/itex] (i.e. the limit is divergent)

    A2. [itex]\stackrel{limit}{_{x→0}}\frac{x-1}{x}=-∞[/itex] (i.e. the limit is divergent)

    A3. [itex]\stackrel{limit}{_{x→0}}\frac{1-2x}{x}+\frac{x-1}{x}=\frac{1-2x+x-1}{x}=\frac{-x}{x}=-1[/itex] (i.e. the limit is convergent)

    I am just confused by the apparent fact that sum of two divergent limits can be a convergent limit. Even though this is apparent in my problem, I still want to make sure if I am not messing up anywhere.
  2. jcsd
  3. Dec 15, 2013 #2

    Ray Vickson

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    You cannot verify that "result" because it is just false, as your example shows. There is a somewhat similar true result, but it involves some extra hypotheses that your example fails to satisfy.
  4. Dec 17, 2013 #3
    you're not summing two divergent limits. You're summing two functions whose individual limits at the same point are both divergent.
  5. Dec 17, 2013 #4


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    What the theorem states is that if both limits on the right side exist, then the limit on the left exists and equals the sum of the limits on the right. That's all.
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