Question

$\sqrt{x} x^{2} \geq \neq \lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx \leq \leq \neq \alpha x_{123} \sqrt[n]{x} \sqrt{x} x^{2} \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right. \leq \geq$

Answer 1

Step-by-step explanation:

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