Question

Помогите решить пожалуйста

Answer 1

$1); ; limlimits _{x to 0}frac{2x}{sqrt{4+x}-sqrt{4-x}}=limlimits _{x to 0}frac{2x(sqrt{4+x}+sqrt{4-x})}{4+x-(4-x)}=limlimits_{x to 0}(sqrt{4+x}+sqrt{4-x})=\\=sqrt{4+0}+sqrt{4-0}=2+2=4\\2); ; limlimits _{x to 0}frac{2-sqrt{x}}{3-sqrt{2x+1}}=frac{2-0}{3-sqrt1}=frac{2}{2}=1$

$3); ; limlimits _{x to 8}frac{sqrt[3]{x}-2}{sqrt{x}-2sqrt2}=limlimits _{x to 8}frac{sqrt[3]{x}-2}{sqrt{x}-sqrt8}=limlimits _{x to 8}frac{(sqrt[3]{x}-2)(sqrt[3]{x^2}+2sqrt[3]{x}+4)(sqrt{x}+sqrt8)}{(sqrt{x}-sqrt8)(sqrt{x}+sqrt8)(sqrt[3]{x^2}+2sqrt[3]{x}+4)}=\\=limlimits_{x to 8}frac{(x-8)(sqrt{x}+sqrt8)}{(x-8)(sqrt[3]{x^2}+2sqrt[3]{x}+4)}=frac{sqrt8+sqrt8}{sqrt[3]{8^2}+2sqrt[3]8+4}=frac{4sqrt2}{4+4+4}=frac{sqrt2}{3}$