Can I use L'Hopital's to show $\lim_{x\to1^-}(1-x)[\frac{d}{dx}(1-x)\sum_{n=1}^\infty a_nx^n]=0$ for $a_n$ a bounded sequence of reals?

My "favourite" counterexample works again. Consider $a_n=(-1)^k$ for $2^k\leqslant n<2^{k+1}$, $k\geqslant 0$.

Then, for $f(x):=\sum_{n=1}^\infty a_n x^n$, we get $g(x):=(1-x)f(x)=x+2\sum_{k=1}^\infty(-1)^k x^{2^k}$. Now let $$h(x)=g(x)+G(\log x),\qquad G(t)=\sum_{n=1}^\infty\frac{2^n-1}{2^n+1}\frac{t^n}{n!}.$$ Then it is easy to check that $h(x)=-h(x^2)$. That is, the function $H(t)=h(e^{-2^{-t}})$ (defined for all real values of $t$) is periodic: $H(t)=H(t+2)$. It is nonconstant, and in fact the linked answer shows that $$H(t)=\frac{2}{\log 2}\sum_{n\in\mathbb{Z}}\Gamma\left(\frac{2n+1}{\log 2}i\pi\right)e^{(2n+1)i\pi t}.$$

Gathering it all, we get $(1-x)f(x)=H\big(-\log_2(-\log x)\big)-G(\log x)$ and $$(1-x)\frac{d}{dx}\big((1-x)f(x)\big)=-\frac{1-x}{x\log x\log 2}H'\big(-\log_2(-\log x)\big)-\frac{1-x}{x}G'(\log x).$$ At $x\to1^-$, the second term vanishes, but the first one oscillates, since $\frac{1-x}{\log x}$ tends to $-1[{}\neq 0]$.