4. [4] Let $a, b$ be constants such that $\lim _{x \rightarrow 1} \frac{(\ln (2-x))^{2}}{x^{2}+a x+b}=1$. Determine the pair $(a, b)$.
Answer: $\quad(-2,1)$ When $x=1$, the numerator is 0 , so the denominator must be zero as well, so $1+a+b=0$. Using l'Hôpital's rule, we must have
$$
1=\lim _{x \rightarrow 1} \frac{(\ln (2-x))^{2}}{x^{2}+a x+b}=\lim _{x \rightarrow 1} \frac{2 \ln (2-x)}{(x-2)(2 x+a)},
$$
and by the same argument we find that $2+a=0$. Thus, $a=-2$ and $b=1$. This is indeed a solution, as can be seen by finishing the computation.