Limit Proof with Rational Functions

1. Aug 30, 2012

Villyer

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. Aug 30, 2012

LCKurtz

It looks fine to me. You should probably observe that for every $a$ in domain $r$, $g(a)\ne 0$.

3. Aug 30, 2012

Villyer

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook