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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?

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Hint: Observe \begin{align} |f(x)(x-x_0)|\leq |f(x_0)||x-x_0| \end{align} for all $x\leq x_0$.

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

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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, $|f(x)(x-x_0)|\leq M|x-x_0|<M\delta\leq\varepsilon$.

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