Question

$\lim_{n \to \infty} a_n \alpha \sqrt{x} \geq \geq \sqrt{x} \sqrt{x} \sqrt[n]{x} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \geq \\ \leq \left \{ {{y=2} \atop {x=2}} \right. \geq x^{2} \sqrt{x} =8$

Answer 1

Step-by-step explanation:

sorry sorry sorry sorry sorry sorry sorry sorry

Answer 2

Step-by-step explanation:

This is your answer

thank you