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Limit Proof with Rational Functions

  • Thread starter Villyer
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Homework Statement



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)##.

Homework 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##

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?
 
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Answers and Replies

  • #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##.
 
  • #3
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I amended the third line of the proof to incorporate that. Thank you!
 

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