The quadratic polynomial $p(x)$ has roots $1$ and $\alpha$, while quadratic polynomial $q(x)$ has roots $1$ and $\beta$. Let $\alpha$ and $\beta$ be the roots of $r(x) = p(x) + q(x)$. Then $\lim_{x\to\infty}\left[\sqrt{p(x)} - \sqrt{q(x)}\right] =$

Let the quadratic polynomials be $p(x) = a(x-1)(x-\alpha)$ and $q(x) = b(x-1)(x-\beta)$.

Given $r(x) = p(x) + q(x)$, we can evaluate $r(x)$ at $x = 1$:
$r(1) = p(1) + q(1) = 0 + 0 = 0$

Since the roots of $r(x)$ are given as $\alpha$ and $\beta$, and $r(1) = 0$, one of the roots must be $1$. Without loss of generality, let $\alpha = 1$.

Then $p(x) = a(x-1)^2$.
Since the roots of $r(x)$ are $1$ and $\beta$, we can write $r(x) = (a+b)(x-1)(x-\beta)$.

Substitute $p(x)$, $q(x)$, and $r(x)$ into $r(x) = p(x) + q(x)$:
$(a+b)(x-1)(x-\beta) = a(x-1)^2 + b(x-1)(x-\beta)$

Dividing by $(x-1)$ for $x \neq 1$:
$(a+b)(x-\beta) = a(x-1) + b(x-\beta)$
$ax - a\beta + bx - b\beta = ax - a + bx - b\beta$
$-a\beta = -a \Rightarrow a(\beta - 1) = 0$

Since $p(x)$ is a quadratic polynomial, its leading coefficient $a \neq 0$, which gives $\beta = 1$.
Thus, $\alpha = \beta = 1$.

This implies $p(x) = a(x-1)^2$ and $q(x) = b(x-1)^2$.

We are required to find:
$\lim_{x\to\infty} \left[\sqrt{p(x)} - \sqrt{q(x)}\right] = \lim_{x\to\infty} \left[\sqrt{a(x-1)^2} - \sqrt{b(x-1)^2}\right] = \lim_{x\to\infty} (\sqrt{a} - \sqrt{b})(x-1)$

For this limit to exist and be a finite value (as per the given options), we must have $a = b$.
Therefore, $\sqrt{a} - \sqrt{b} = 0$, and the limit evaluates to $0$.

Answer: $0$