<s>[INST] Proving $\lim\limits_{x\to a}(a-x)\int\limits_0^x\frac{f(y)}{(a-y)^2}\,dy=f(a)$ How to prove for a continuous function $f$, the following limit holds? 
$$\lim_{x\to a}\,(a-x)\int_0^x\frac{f(y)}{(a-y)^2}\,dy=f(a)$$
 [/INST] Here is general proof without using limit theorems:
First remark that
$$(x-a) \cdot \int_0^x \frac{f(a)}{(a-y)^2} = f(a)+ \frac{f(a) \cdot (x-a)}{a}$$
hence
$$\left|(a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) \right| \leq \left| (a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} \right| + \underbrace{\left|\frac{f(a) \cdot (x-a)}{a} \right|}_{\to 0 \, (x \to a)}$$
Now we have by the first equation
$$(a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} = (x-a) \cdot \int_0^x \frac{f(y)-f(a)}{(a-y)^2} \, dy $$
Let $\varepsilon>0$. Since $f$ is continuous we find $\delta>0$ such that $|f(y)-f(a)| \leq \varepsilon$ for all $y \in B(a,\delta)$. Thus 
$$\left|(x-a) \cdot \int_{(0,x) \cap B(a,\delta)} \frac{f(y)-f(a)}{(a-y)^2}\right| \leq \left(1- \frac{x-a}{a} \right) \cdot \varepsilon \\ 
\left|(x-a) \cdot \int_{(0,x) \cap B(a,\delta)^c} \frac{f(y)-f(a)}{(a-y)^2}\right| \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta^2} \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta_0^2} $$
for some fixed $\delta_0>0$. Hence 
$$ \left| (a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} \right| \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta_0^2} + \left(1- \frac{x-a}{a} \right) \cdot \varepsilon \to 0 \quad (\varepsilon \to 0, x \to a)$$
</s>