Step 1: Plan:
Use the sum and product of roots on $r(x)$ instead of factoring it. Assume monic quadratics.
Step 2: Write p, q and r:
$p(x) = x^2-(1+\alpha)x+\alpha$ and $q(x) = x^2-(1+\beta)x+\beta$. Adding,
\[ r(x) = 2x^2 - (2+\alpha+\beta)x + (\alpha+\beta) \]
Step 3: Vieta on r:
The roots of $r$ are $\alpha$ and $\beta$, so sum $= \alpha+\beta = \frac{2+\alpha+\beta}{2}$. This gives $\alpha+\beta = 2$.
Product $= \alpha\beta = \frac{\alpha+\beta}{2} = 1$.
So $\alpha+\beta = 2$ and $\alpha\beta = 1$. These are the roots of $t^2-2t+1 = 0$, so $\alpha=\beta=1$.
Step 4: Take the limit:
Both polynomials become $(x-1)^2$, so the difference of their square roots is exactly $0$ for all $x$, and the limit is $0$.
Final Answer:
The two polynomials are identical, so the limit is $0$, option (A).
\[ \boxed{0} \]