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I need some help with a calculus homework question. Here is said question:

Let there be two polynomials $q$ and $p$ such that $\deg(p)\leq\deg(q)+1$ and $q(x)\neq0$ for all $x\in\mathbb{R}$.

Show that that rational function $f:\mathbb{R\rightarrow R}$ that is defined as $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ is uniformly continuous

I'm guessing that they want me to prove that all Rational Functions are uniformly continuous but I have no clue as to how to do that (seeing as throughout my internet search I saw many people referencing it)

Any help is appreciated, Thx!

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    $\begingroup$ Note $f$ is continuous and for large $x$, $f(x)\sim ax+b$. $\endgroup$ – David Mitra Feb 18 '14 at 16:43
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In fact, such a function is Lipschitz continous but the degree assumption is critical here. This can be seen as follows: Let $n = \deg(p)$, $m = \deg (q)$. Since $q(x) \neq 0$, the function $f$ is differentiable for all $x \in \mathbb{R}$. Now, $$ f'(x) = \frac{p'(x)q(x) - p(x)q'(x)}{q(x)^2} $$ is bounded by the degree assumption (the denominator is of degree $2m$ while the numerator is of degree $\max\{(n-1) + m, n + (m-1)\} = n + m - 1 \leq 2m$). Thus $f$ is Lipschitz and in particular uniformly continuous.

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Some hints:

  • As the degree of $p(x)$ is smaller than the degree of $q(x)$ plus one, we have that $p(x)/q(x)$ behaves as $ax+b$ asymptotically.

  • Now consider the function $h(x)=\frac{p(x)}{q(x)}-(ax+b)$, and note that $h(x)$ tends to zero as $x$ tends to plus/minus $\infty$.

  • A continuous function is uniformly continuous on compact sets.

  • The sum of two uniformly continuous functions is uniformly continuous

Do you see how to combine these statements into a proof? (Involves some case-checking).

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  • $\begingroup$ The hypotheses allow, e.g., that $p$ has degree $4$ and $q$ has degree $3$. So the quotient is at worst nearly a linear function for $x$ large. $\endgroup$ – David Mitra Feb 18 '14 at 16:50
  • $\begingroup$ Thank you, I totally misread the question. I have edited the answer accordingly. I hope it is better now. $\endgroup$ – Henrik Feb 18 '14 at 17:07

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