# prove $\frac{f(x)}{x}$ is uniformly continuous in $[1,\infty]$

Let $$f$$ a real function defined in $$[1,\infty]$$ such that $$\frac{f'(x)}{x}$$ is bounded in the domine of $$x$$. Show that function $$\frac{f(x)}{x}$$ is uniformly continuous in $$[1,\infty]$$.Can you weaken hypotheses? I was thinking to use the next theorem: $$Theorem:$$ If $$f'$$ exist and is bounded in $$(-\infty,\infty)$$ then $$f$$ is uniformly continuous in $$(-\infty,\infty)$$. But i don't know how to apply it to resolve this.

Lets us first establish that the bound on $$\frac{f'(x)}{x}$$ for $$x \geq 1$$implies $$|f'(x)|\leq Cx$$. Then we calculate: $$|f(x)|=|\int_1^xf'(t)dt+f(1)|\leq C\int_1^x tdt+|f(1)|=\frac{1}{2}Cx^2+K$$ with $$K=-\frac{C}{2}+|f(1)|$$. Define $$p(x)=\frac{f(x)}{x}$$. Then: $$p'(x)=\frac{f'(x)}{x}-\frac{f(x)}{x^2}$$ The first term is bounded by assumption. The second term is bounded for $$x \geq 1$$ since $$|\frac{f(x)}{x^2}|\leq \frac{\frac{1}{2}Cx^2+K}{x^2} \leq C_2 <\infty$$ Overall, we have $$|p'(x)|\leq C+C_2<\infty$$ and you can use your theorem to conclude.