If E is a bounded set and $p(x)$ is a polynomial that has no real roots, explain why $\frac{1}{p(x)}$ is uniformly continuous on E
Let $f(x)=\frac{1}{p(x)}$, a function $f(x)$ is said to be uniformly continuous on E if
$\forall \epsilon>0$, $\exists \delta>0$, such that $\forall x_0 \in E$, $\lvert x-x_0 \rvert < \delta \implies \lvert f(x)-f(x_0) \rvert < \epsilon$
Let $p(x)=ax^2+bx+c$ for some constants a,b,c
We know that $p(x)$ has no real roots if $b^2-4ac<0$
So $\forall \epsilon>0 $, we need to find $\delta>0$ such that $\forall x_0 \in E$, $\lvert x-x_0 \rvert <\delta \implies \lvert \frac{1}{ax^2+bx+c}-\frac{1}{ax_0^2+bx_0+c} \rvert < \epsilon$
$=\lvert \frac{(ax_0^2+bx_0+c)-(ax^2+bx+c)}{(ax^2+bx+c)(ax_0^2+bx_0+c)} \rvert= \lvert \frac{a(x^2-x_0^2)+b(x-x_0)}{(ax^2+bx+c)(ax_0^2+bx_0+c)} \rvert=\lvert \frac{[a(x-x_0)(x+x_0)]+b(x-x_0)}{(ax^2+bx+c)(ax_0^2+bx_0+c)} \rvert=\lvert \frac{(x-x_0)[a(x+x_0)+b]}{(ax^2+bx+c)(ax_0^2+bx_0+c)}\rvert=\lvert x-x_0 \rvert \lvert \frac{a(x+x_0)+b}{(ax^2+bx+c)(ax_0^2+bx_0+c)} \rvert $
We haven't went over derivatives yet so I can't use them
Claim. Let $p\in \Bbb R[X]$ be a polynomial and assume that $p(x)\ne 0$ for all $x\in \Bbb R$. Then $f(x):=\frac1{p(x)}$ is uniformly continuous on $\Bbb R$.
Proof. The case of constant $p$ is trivial, hence if we write $p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_0$ with $a_n\ne 0$ we may assume $n>0$. Also, without loss of generality $a_n>0$. Then for $x\ne 0$ $$p(x)=x^n\cdot\left(a_n+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\ldots+\frac{a_0}{x^n}\right) $$ Let $h=\max\{|a_0|,|a_1|,\ldots,|a_{n-1}|\}$. For sufficiently large $x$, say for $|x|>\max\left\{1,\frac{(n+1)h}{a_n}\right\}$, we have $$\left|\frac{a_{n-k}}{x^k}\right|\le\left|\frac{a_{n-k}}{x}\right|<\left|\frac{a_na_{n-k}}{(n+1)h}\right|\le \frac{a_n}{n+1}$$ and therefore $|p(x)|>|x^n|\cdot\left(a_n-n\cdot\frac{a_n}{n+1}\right)=\frac{a_n}{n+1}|x|^n\ge\frac{a_n}{n+1}|x|$.
Let $\epsilon>0$ be given. As just sseen, we have $\left|\frac1{p(x)}\right|<\frac{n+1}{a_nx}<\frac\epsilon2$ for $|x|$ large enough, in fact for $|x|>C_\epsilon$ with $C_\epsilon=\max\left\{1,\frac{(n+1)h}{a_n},\frac{2(n+1)}{a_n\epsilon}\right\}$. As $x\mapsto \frac1{p(x)}$ is continuous on the compact interval $[-C_\epsilon-1,C_\epsilon+1]$ it is uniformly continuous there. Hence for some $\delta>0$ (wlog. $\delta<1$) we have that $|f(x)-f(y)|<\epsilon$ for all $x,y$ with $x,y\in[-C_\epsilon-1,C_\epsilon+1]$ with $|x-y|<\delta$. If we consider arbitrary $x,y\in\Bbb R$ with $|x-y|<\delta$, then both are in $[-C_\epsilon-1,C_\epsilon+1]$ or (as $\delta<1$) both are in $\Bbb R\setminus[-C_\epsilon,C_\epsilon]$; in both cases we find $|f(x)-f(y)|<\epsilon$. $_\square$