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Continuity of a Rational Function at a number help

  1. Aug 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Find continuity of function f(x)= (x^2-1)/(x-1) at x = 1

    2. Relevant equations

    limit f(x) as x-> = L

    3. The attempt at a solution

    i KNOW it can be easily solved by stating that at x = 1 function becomes infinity,so discontinous it is actually...But as we do in finding domain and range of rational functions,we make an altered functions by factorization...in this case by factorization f(x) = x+1 which when x = 1 give limit = 2....I couldn't understand the notion here why we don't factorize here?? please tell me thoroughly in layman's terms
     
    Last edited: Aug 16, 2013
  2. jcsd
  3. Aug 16, 2013 #2

    CAF123

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    Gold Member

    You have the limit right. Only for ##x \neq 1,## does $$\frac{x^2 - 1}{x-1} = x + 1$$

    Since you are not interested in what happens to the function at x=1 (even though you know it is discontinuous and the value of f(1) does not exist) you may factorise to find the limit as you have done.
     
  4. Aug 16, 2013 #3

    HallsofIvy

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    The answer to your question, whether [itex](x^2- 1)/(x- 1)[/itex] is continuous at x= 1 or not is, clearly, that is it NOT continuous there. It is not even defined at x= 1! (However, the function does NOT "become infinity" at x= 1 nor does it "go to infinity" as x approaches 1.)

    IF the question were, instead, whether the function defined as "[itex]f(x)= (x^2- 1)/(x- 1)[/itex] if x is NOT equal to 1, f(1)= 2" is continuous at x= 1, then the answer is "yes".
     
  5. Aug 16, 2013 #4
    but if we factorize it the limit becomes 2...
     
  6. Aug 16, 2013 #5

    Ray Vickson

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    The function that you wrote, which is
    [tex] f(x) = x^2 - \frac{1}{x} -1[/tex]
    is obviously finite and continuous at x=1.

    I interpreted the question exactly as written and when read/parsed by standard rules. If you had meant
    [tex] f(x) = \frac{x^2-1}{x-1}[/tex]
    you would have know to use brackets, like this: f(x) = (x^2-1)/(x-1).
     
  7. Aug 16, 2013 #6
    pardon my brain-fart sire
     
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