Let's determine all the number pairs $(A, B)$ for which there exist sequences $a_{n}, b_{n}$ such that $\lim \left(a_{n} b_{n}\right) = A$ and $\lim \left(a_{n} + b_{n}\right) = B$. For which $A$ and $B$ will it be true that the sequences $a_{n}$ and $b_{n}$ themselves are convergent?
Given the sequences $a_{n}$ and $b_{n}$, consider the sequence $d_{n}=\left(a_{n}+b_{n}\right)^{2}-4 a_{n} b_{n}=\left(a_{n}-b_{n}\right)^{2}$. If the sequences $\left(a_{n}+b_{n}\right)$ and $\left(a_{n} b_{n}\right)$ are convergent, then $\left(d_{n}\right)$ is also convergent, and by the elementary relationships of limit calculations, $\lim _{n \rightarrow \infty} d_{n}=$ $\lim _{n \rightarrow \infty}\left[\left(a_{n}+b_{n}\right)^{2}-4 a_{n} b_{n}\right]=B^{2}-4 A$. The elements of $\left(d_{n}\right)$ are non-negative numbers, so the limit of the sequence cannot be negative. The pair $(A, B)$ thus satisfies the problem's requirements if and only if $B^{2} \geqq 4 A$.

We will show that the found condition is sufficient: if $B^{2} \geqq 4 A$, then there exist sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ such that their product converges to $A$ and their sum converges to $B$.

Consider the numbers $\alpha=\frac{B+\sqrt{B^{2}-4 A}}{2}$ and $\beta=\frac{B-\sqrt{B^{2}-4 A}}{2}$. Since $\alpha \beta=A$ and $\alpha+\beta=B$, the sequences $a_{n}=\alpha$ and $b_{n}=\beta$ clearly satisfy the conditions.

It is also clear that if we write $\left\{a_{n}, b_{n}\right\}=\{\alpha, \beta\}$ for all $n$, we will obtain such sequences. This means that if $\alpha \neq \beta$, i.e., $B^{2}>4 A$, then there exist sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ such that $\lim _{n \rightarrow \infty} a_{n} b_{n}=A$ and $\lim _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=B$, although the sequences themselves are not convergent. Such sequences can be obtained, for example, by swapping every second element of the above constant sequences, i.e., for even $n$, $a_{n}=\alpha$ and $b_{n}=\beta$, and for odd $n$, $a_{n}=\beta$ and $b_{n}=\alpha$. The answer to the second part of the problem is therefore negative in the case $B^{2}>4 A$.

If $B^{2}=4 A$, then the limit of the sequence $\left(d_{n}\right)$ examined in the first part of the solution is 0, so $\lim _{n \rightarrow \infty}\left(a_{n}-b_{n}\right)=0$ is also true. Thus, $a_{n}=\frac{a_{n}+b_{n}}{2}+\frac{a_{n}-b_{n}}{2}$ is convergent as the sum of convergent sequences, and similarly, $b_{n}=\frac{a_{n}+b_{n}}{2}-\frac{a_{n}-b_{n}}{2}$ is also convergent. In this case, clearly $\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}=\frac{B}{2}$.

Remarks: 1. The solution can also be formulated differently. Since the system of equations $a_{n} b_{n}=u_{n}, a_{n}+b_{n}=v_{n}$ (where $a_{n}$ and $b_{n}$ are the unknowns) has a solution for every $n$, the discriminant of the quadratic equation, $v_{n}^{2}-4 u_{n} \geqq 0$, can be written based on the relationships between the roots and coefficients. Then this inequality also holds for the limit of the sequence: $B^{2}-4 A \geqq 0$.

2. In the case $B^{2} \geqq 4 A$, not only the sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ given in the solution are suitable. Starting from any sequences $\alpha_{n} \rightarrow \alpha$ and $\beta_{n} \rightarrow \beta$, the condition is satisfied for any sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ such that $\left\{a_{n}, b_{n}\right\}=\{\alpha_n, \beta_n\}$. Generally, we can only state that every sequence $\left(a_{n}\right)$ and $\left(b_{n}\right)$ that satisfies the condition has a limit point, and the limit points of their union are $\alpha$ and $\beta$.