Sagot :
Answer:
[tex]\sqrt[n]{x} \sqrt{x} \sqrt{x} \pi x^{2} \\ \neq \alpha \beta \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right][/tex]
Explanation: