Limit of a certain expression

The statement is true! In order to prove this, we recall following modifacted Taylor-forumla: We have $$f(t) = f(x) + f'(x)(t-x) + \int_x^t (f'(s)-f'(x)) \,\mathrm{d}s$$ for $x,t\in (0,1)$. This formula can be easily checked by integration. Note that we only need that $f$ is continuously differentiable on $(0,1)$.

We apply this formula to $t=(1+\varepsilon)x$ in order to find $$f'(x) x = \frac{1}{ε} \left( f((1+ε)x)-f(x) - \int_x^{(1+ε)x} (f'(s)-f'(x)) \,\mathrm{d}s \right).$$ Now, we know that $f((1+ε)x)-f(x) \rightarrow 0$ for $x \rightarrow 0+$ and fixed $ε >0$. Moreover, we can bound the integrand using$$|f'(s)-f'(x)| \leq |s-x| \sup_{h \in [x,(1+ε)x]} |f''(h)| \leq |s-x| · Cx^{-2} \leq Cεx^{-1}.$$ Thus $$ \frac{1}{ε}\left|\int_x^{(1+ε)x} (f'(s)-f'(x)) \,\mathrm{d}s \right| \leq \int_x^{(1+ε)x} C x^{-1} \,\mathrm{d}s \leq Cε.$$ All togehter implies $$\limsup_{x \rightarrow 0+} |f'(x) x| \leq Cε$$ and, since $ε>0$ is arbitary, we get already the claim.

Inspecting the proof, we see that we only used $\lim\limits_{x \rightarrow 0+} f(x) =0$ and $|f''(x)| \leq C x^{-2}$, and that $f$ is twice differentiable.

Edit: My initial formula was wrong. I have repaired the proof.