$f$ defined on $[1,\infty )$ is uniformly continuous. Then $\exists M>0$ s.t. $\frac{|f(x)|}{x}\le M$ for $x\ge 1$. $f$ defined on $[1,\infty )$ is uniformly continuous. Then $\exists M>0$ s.t. $\frac{|f(x)|}{x}\le M$ for $x\ge 1$.
I know f uniformly continuous $\implies \forall\varepsilon > 0s.t.\forall x,y\in [1,\infty)$,
$$|f(x)-f(y)|<\varepsilon \space\space\forall |x-y|<\delta$$
By Mean value theorem, $\exists c\in (x,y)\forall x,y\in [1,\infty)$ such that 
$$\frac{f(x)-f(y)}{x-y}=f'(c)$$
$$\frac{|f(x)-f(y)|}{|x-y|}=|f'(c)|<M$$ for some $M>0$
$$|f(x)-f(y)|<M|x-y|<M\delta $$
How can I get an expression for $|f(x)|/x$?
 A: Assume that $f(1)=0$ for the moment. Let $\epsilon=1$. Then there is $\delta $ such that 
$$|f(x) - f(y)| < 1$$
whenever $|x-y|\leq\delta$. Let $x\in [1, \infty)$. The idea is that you chop the interval $[1, x]$ into small intervals so that you can apply the above inequality. Now there is $n\in \mathbb N$ such that 
$$1 + n\delta \leq x < 1 + (n+1)\delta,$$
(Just chopping the intervals) Then write $x_m = 1+ m\delta$, 
$$\begin{split} |f(x)| &= |f(x)-f(1)| \\
&\leq |f(x) - f(x_{n})| +  |f(x_{n}) - f(x_{n-1})| + \cdots \\
&\ \ \ +|f(x_2) - f(x_1)|+ |f(x_1) - f(1)| \\
&\leq 1 + 1 + \cdots + 1 = n+1 <2n\\
&< \frac{2}{\delta}(1+ \delta n)\\
&\leq Mx\ .
\end{split}$$
By letting $M = \frac{2}{\delta}$ (note that $\delta $ is fixed). Thus 
$$\frac{|f(x)|}{x} \leq M\ .$$
In general, if $f(x) \neq 0$, then letting $g(x) = f(x) - f(1)$, then $g$ is also uniformly continuous. Thus there is $M$ such that 
$$\frac{|f(x) - f(1)|}{x} = \frac{|g(x)|}{x} \leq M$$
which implies 
$$\frac{|f(x)|}{x} \leq M+ \frac{|f(1)|}{x} \leq M + |f(1)|\ .$$
