Question: Given that $f$ is uniformly continuous on $[0,\infty)$, show that $\lim_{x \to \infty} \frac{f(x)}{1+x^2} = 0$.
What I know:
Uniform continuity: For every $\epsilon > 0$, $\exists\,\delta > 0$ such that for all $x,y \in [0,\infty)$, if $|x-y| < \delta$, then $|f(x)-f(y)| < \epsilon$.
Want to show:
$|\frac{f(x)}{1+x^2}| < \epsilon$ as $x \to \infty$.
Attempt #1: I attempted to show that $f$ is Lipschitz but doesn't seem to be the case.
Attempt #2: Tried to re-write $|x-y| < \delta$ as $|x| - |y| < \delta$ and $|f(x)-f(y)| < \epsilon$ as $|f(x)| - |f(y)| < \epsilon$ to manipulate the expression but to no avail.
Also thought of trying to show that $f(x)$ "moves slower" than $1+x^2$ or $x^2$ in the limit but not sure how to put it exactly.
Any help will be appreciated. Thanks!
Since $f$ is uniformly continuous, then for $\varepsilon=1$, there exists a $\delta>0$, such that $$ |x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)|<1. $$ Pick $h$, with $0<h<\delta$. So, for every $x>0$, set $n=\big[\frac{x}{h}\big]$. Then $$ |f(x)-f(0)|\le |f(0)-f(h)|+|f(h)-f(2h)|+\cdots+\\|f\big((n-1)h\big)-f(nh)|+|f(nh)-f(x)| < n+1<\frac{x}{h}+1, $$ and hence $$ \frac{|f(x)|}{1+x^2}< \frac{|f(0)|+\frac{x}{h}+1}{x^2+1}\to 0 $$ as $x\to\infty$.