Question

Помогите с пределами. Даю 80 баллов

Answer 1

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

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

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

Answer 2

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