$f\in C^2((0,1)),\lim_{x\to 1^{-}}f(x)=0,\exists C\forall x\in (0,1)$,$(1-x)^2|f''(x)|\leqslant C$,then $\lim_{x\to 1^{-}} (1-x)f'(x)=0$ Question:
Suppose that $f\in C^2((0,1))$,$\lim\limits_{x\to 1^{-}}f(x)=0$.
Assume that there exists a constant $C>0$ such that
$\forall x\in (0,1)$, $(1-x)^2|f''(x)|\leqslant C$.
Prove that $\lim_\limits{x\to 1^{-}} (1-x)f'(x)=0$.
Attempt:
I've tried to use Taylor expansion and got
$0=f(x)+f'(x)(1-x)+\frac{f''(c)}{2}(1-x)^2,c\in (x,1).$
But that's not enough.

ParamanandSingh's hint helped enourmously.
Given $\epsilon>0$, there exist $\delta \in (0,1)$ such that if $x > 1 - \delta$, then $|f(x)| < \epsilon$.
For any $0 < x_0 < x < \tfrac12(1+x_0)$, we have $|f''(x)| \le C/(1-x)^2 \le 4C/(1-x_0)^2$.  Hence
$$ f'(x) = f'(x_0) + \int_{x_0}^x f''(y) \, dy \ge f'(x_0) - 4C\frac{x-x_0}{(1-x_0)^2} .$$
Similarly,
$$ f'(x) = f'(x_0) + \int_{x_0}^x f''(y) \, dy \le f'(x_0) + 4C\frac{x-x_0}{(1-x_0)^2} .$$
Now suppose $f'(x_0) \ge L /(1-x_0)$ for some $x_0>1-\delta$, where $L$ will be chosen later.  Then for
$$ x_0 \le x \le x_1 := \min\left\{\frac{1+x_0}2, x_0 + \frac{L(1-x_0)}{8C}\right\} $$
we have
$$ f'(x) \ge \frac L{2(1-x_0)} $$
Hence if $L \le 4C$, we have
$$ 2 \epsilon > f(x_1) - f(x_0) = \int_{x_0}^{x_1} f'(x) \, dy \ge \frac L{2(1-x_0)} \cdot \frac{L(1-x_0)}{8C}  = \frac{L^2}{16C} .$$
Without loss of generality, we can assume $32 \epsilon < 16C^2$.
Hence if we set $ L^2 \in( 32 \epsilon, 16C^2] $, we obtain a contradiction.
Similarly, if $f'(x_0) \le -L/(1-x_0)$, we obtain a similar contradiction.
Hence for any $x_0 > 1 - \delta$, as long as $\epsilon < C^2/2$, we have
$$ (1-x_0)|f(x_0)| \le \sqrt{32 \epsilon} .$$
