Q:

Limits - prove or disprove

If $\lim_{x \to 0^+} f(x) = 0$ and $(\forall x>0)( \exists 0<c_x<x)$ s. t. $f(c_x)>f(x)$, and $\forall x>0, f(x)>0$ do we have a contradiction?
I tried to build a sequence of x values that approaches $0$ but its f values form an ascending sequence but failed to show that it is actually ascending.

A:

If $f$ is allowed to be discontinuous in every right neighbourhood of $0$, then the condition does not prevent $\lim_{x\to 0^+}f(x)=0$. For instance, consider the function $f:(0,1]\to\Bbb R$ such that $f(x)=\frac1{n-1}+\frac1n-x$ for all $n\in\Bbb N$, $n\ge2$ and for all $x$ such that $\frac1n<x\le \frac1{n-1}$. Namely, $$f(x)=\left\lfloor x^{-1}\right\rfloor^{-1}+\left(\left\lfloor x^{-1}\right\rfloor+1\right)^{-1}-x$$

On the other hand, if there is some $\varepsilon>0$ such that $\left.f\right\rvert_{(0,\varepsilon)}$ is continuous, then yes, that condition prevents $\lim_{x\to 0^+}f(x)=0$. In fact, consider $0<x_0<\varepsilon$ and let $\delta=\inf\{\alpha>0\,:\, f(\alpha)\ge f(x_0)\}$. If $\delta>0$, then by continuity $f(\delta)\ge f(x_0)$ and, therefore, there must be some $0<c_\delta<\delta$ such that $f(c_\delta)\ge f(\delta)\ge f(x_0)$, against $\delta$ being the greatest lower bound. Therefore $\delta=0$, but then $\limsup_{x\to 0^+}f(x)\ge f(x_0)> 0$.

