помогите решить очень надо. Created by Rikudo. algebra-ru

помогите решить очень надо...

Rikudo @Rikudo

August 2022 1 8 Report

помогите решить очень надо

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NNNLLL54

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$1)\quad \lim\limits _{x \to \infty} \frac{3x^3-x^2+2x}{x^3-x^4} = \lim\limits _{x \to \infty} \frac{x(3x^2-x+2)}{x(x^2-x^3)} = \lim\limits _{x \to \infty} \frac{3x^2-x+2}{x^2-x^3} =\\\\= \lim\limits _{x \to \infty} \frac{\frac{3}{x}-\frac{1}{x^2}+\frac{2}{x^3}}{\frac{1}{x}-1} =[\, \frac{0}{-1} \, ]=0$

$2)\quad \lim\limits _{x \to \infty} \frac{\sqrt{4x^2+2x}-\sqrt{x+3}}{\sqrt[3]{64x^3+1}+2} =[\, \frac{:x}{:x} \, ]= \lim\limits _{x \to \infty} \frac{\sqrt{4+\frac{2}{x}}-\sqrt{\frac{1}{x}+\frac{3}{x^2}}}{\sqrt[3]{64+\frac{1}{x^3}}+\frac{2}{x}} =\\\\=\frac{\sqrt{4+0}-\sqrt{0+0}}{\sqrt[3]{64+0}+0} = \frac{2}{4} = \frac{1}{2}$

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