Question

Помогите с пределами. Даю 80 баллов

Answer 1

$1)\; \; \lim\limits _{x \to \infty}\frac{(1-x)^4-(1+x)^4}{(1+x)^3+(x-1)^3}=\\\\=\lim\limits _{x \to \infty}\frac{x^4-4x^3+6x^2-4x+1-(x^4+4x^3+6x^2+4x+1)}{1+3x+3x^2+x^3+x^3-3x^2+3x-1}=\\\\=\lim\limits _{x \to \infty}\frac{-8x^3-8x}{2x^3+6x}=\lim\limits _{x \to \infty}\frac{-8x^3}{2x^3}=-4$

$2)\; \; \lim\limits _{x \to \infty}\; \frac{\sqrt[3]{x}-9x^2}{3x-\sqrt[4]{9x^8+1}}=\Big [\frac{:x^2}{:x^2}\Big ]=\lim\limits _{x \to \infty}\; \frac{\frac{1}{x^{5/3}}-9}{\frac{3}{x}-\sqrt[4]{9+\frac{1}{x^2}}}=\frac{0-9}{0-\sqrt[4]9}=\\\\=\sqrt[4]{9^3}=\sqrt[4]{3^6}=3\sqrt[4]{3^2}=3\sqrt3$

$3)\; \; \lim\limits _{x \to \infty}\; x(\sqrt{x^2+1}-\sqrt{x^2-1})=\lim\limits _{x \to \infty}\; \frac{x(x^2+1-(x^2-1))}{\sqrt{x^2+1}+\sqrt{x^2-1}}=\\\\=\lim\limits _{x \to \infty}\; \frac{2x}{\sqrt{x^2+1}+\sqrt{x^2-1}}=\Big [\frac{:x}{:x}\Big ]=\lim\limits _{x \to \infty}\frac{2}{\sqrt{1+\frac{1}{x^2}}+\sqrt{1-\frac{1}{x^2}}}=\frac{2}{1+1}=1$

Answer 2

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