Show $\lim_{x \to x_0^+} f(x)(x-x_0) =0$ when $f(\mathbb{R}) \subset \mathbb{R}^+$ & monotone increasing. 
Show $\lim_{x \to x_0^+} f(x)(x-x_0) =0$ when $f(\mathbb{R}) \subset \mathbb{R}^+$ & monotone increasing. 


Try
I need to show,
$$
\forall \epsilon >0, \exists \delta >0 : x \in (x_0, x_0 + \delta) \Rightarrow |f(x) (x-x_0)| < \epsilon
$$
I think I could find some upper bound $M >0$ such that $|f(x) (x-x_0)| \le M |x - x_0|$. 
Let $M = f(x_0 + \epsilon)$, and let $\delta = \frac{\epsilon}{\max \{2M, 2 \}}$, then clearly $f(x) \le f(x_0 + \epsilon) = M$
But I'm not sure $|f(x) (x-x_0)| \le M |x - x_0|$. 
Any hint about how I should proceed?

Use $M=f(x_0+1)$ and cosider $\delta=\min\{\frac{1}{2},\frac{\epsilon}{2M}\}$.
