# 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?

Hint: Observe \begin{align} |f(x)(x-x_0)|\leq |f(x_0)||x-x_0| \end{align} for all $$x\leq x_0$$.
Use $$M=f(x_0+1)$$ and cosider $$\delta=\min\{\frac{1}{2},\frac{\epsilon}{2M}\}$$.
Fix $$\varepsilon>0$$. Let $$M=f(x_0+1)$$ and choose $$\delta=\mathrm{min}\{1,\frac{\varepsilon}{M}\}$$. For each $$x\in(x_0,x_0+\delta)$$, $$|f(x)|\leq M$$ since $$f$$ is strictly increasing. Thus, $$|f(x)(x-x_0)|\leq M|x-x_0|.