Question

rasgrahan kara in marathi ​

Answer 1

Answer:

ωω$\sqrt{x} x^{2}$

Explanation:

$\lim_{n \to \infty} a_n x^{2} \sqrt{x}$

$\lim_{n \to \infty} a_n \neq \sqrt{x}$$\pi \sqrt{x} x^{2}$

$\beta x_{123} \frac{x}{y} \alpha$