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Limit Proof with Rational Functions

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data

    If r is a rational function, use Exercise 57 to show that ##\mathop {\lim }\limits_{x \to a} \space r(x) = r(a)## for every number a in the domain of r.

    Exercise 57 in this book is: if p is a polynomial, show that ##\mathop {\lim }\limits_{x \to a} \space p(x) = p(a)##.

    2. Relevant equations

    Limit Laws #5 (in this textbook): ##\mathop {\lim }\limits_{x \to a} \frac{f(x)}{g(x)} = \frac{\mathop {\lim }\limits_{x \to a} \space f(x)}{\mathop {\lim }\limits_{x \to a} \space g(x)} \space if \space \mathop {\lim }\limits_{x \to a} \space g(x) ≠ 0##

    3. The attempt at a solution

    A rational function is defined as ##\frac{p(x)}{g(x)}## where ##p(x)## and ##g(x)## are polynomials, so ##r(x)## can be written as ##\frac{p(x)}{g(x)}##.
    Therefore ##\mathop {\lim }\limits_{x \to a} \space r(x) = \mathop {\lim }\limits_{x \to a} \space \frac{p(x)}{g(x)}## where ##p(x)## and ##g(x)## are polynomials.
    By the limit laws, ##\mathop {\lim }\limits_{x \to a} \space \frac{p(x)}{g(x)} = \frac{\mathop {\lim }\limits_{x \to a} \space p(x)}{\mathop {\lim }\limits_{x \to a} \space g(x)} \space if \space \mathop {\lim }\limits_{x \to a} \space g(x) ≠ 0##, which is true for all a in the domain of r.
    From Exercise 57, ##\mathop {\lim }\limits_{x \to a} \space p(x) = p(a)## and ##\mathop {\lim }\limits_{x \to a} \space g(x) = g(a)## (since ##p(x)## and ##g(x)## are both polynomials).
    So ##\mathop {\lim }\limits_{x \to a} \space r(x) = \frac{p(a)}{g(a)} = r(a)## for every a in the domain of r.

    I'm worried that I didn't address the possibility of a not being in the domain of r enough, and I don't have confidence in my wording. Is what I have okay?
     
    Last edited: Aug 30, 2012
  2. jcsd
  3. Aug 30, 2012 #2

    LCKurtz

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    It looks fine to me. You should probably observe that for every ##a## in domain ##r##, ##g(a)\ne 0##.
     
  4. Aug 30, 2012 #3
    I amended the third line of the proof to incorporate that. Thank you!
     
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