Prove or disprove that if $\lim\limits_{x\to0^+}f(x)=0$ and $|x^2f''(x)|\leq c$ then $\lim\limits_{x\to0^+}xf'(x)=0$
For $0 < y < x < 1$, by Taylor's theorem there exists $\theta \in (0,1)$ such that
$$f(y) = f(x) + f'(x)(y-x) + \frac{1}{2} f''(x - \theta(x-y)) (y-x)^2$$
Taking $y = (1-\eta) x$ where $0 < \eta < 1/2$ we have
$$f(y) - f(x) = -\eta xf'(x) +\frac{\eta^2}{2}x^2f''(x(1 -\theta\eta)) $$
and since $y \to 0+$ as $x \to 0+$,
$$0 = \lim_{x \to 0+}\frac{f(y) - f(x)}{\eta} = \lim_{x \to 0+}\left(-xf'(x)+ \frac{\eta}{2}\frac{1}{(1 - \theta\eta)^2} [x(1-\theta\eta)]^2 f''(x(1 -\theta\eta)) \right) $$
Since the limit on the RHS is $0$, for any $\epsilon > 0$ if $0 < x < \delta$ we have
$$\tag{*}|x f'(x)| \leqslant \epsilon + \frac{\eta}{2(1 - \theta\eta)^2} [x(1-\theta\eta)]^2 |f''(x(1 -\theta\eta))| \leqslant \epsilon + \frac{\eta}{2(1 - \theta\eta)^2}C,$$
Since $0 < \eta < 1/2$, we have
$$\frac{\eta}{(1- \theta \eta)^2} < \frac{\eta}{(1 - \theta/2)^2} < 4\eta,$$
and the last term on the RHS of (*) can be made arbitrarily small.
It follows that
$$\lim_{x \to 0+} x f'(x) = 0$$