Question

Кто решит, тот будет просто самый крутой математик)
Заранее огромное спасибо)

Answer 1

$1)quad limlimits _{n to infty} frac{sqrt{9n^2}+2n+1}{sqrt[3]{n^6}+1}= limlimits _{n to infty} frac{3n+2n+1}{n^2+1} = limlimits _{n to infty} frac{5n+1}{n^2+1} =\\=limlimits _{n to infty} frac{n^2(frac{5}{n}+frac{1}{n})}{n^2(1+frac{1}{n^2})} =frac{0}{1}=0\\2)quad limlimits _{x to 1} frac{x^2-sqrt{x}}{sqrt{x}-1}=limlimits _{x to 1} frac{sqrt{x}((sqrt{x})^3-1)}{sqrt{x}-1} =limlimits_{x to 1}frac{sqrt{x}(sqrt{x}-1)(x+sqrt{x}+1)}{sqrt{x}-1} =$

$=limlimits _{x to 1} (sqrt{x}(x+sqrt{x}+1))=1cdot (1+1+1)=3\\3)quad limlimits _{x to infty} (sqrt{x^2-2x-4}-sqrt{x^2-7x+5})=\\ =limlimits _{x to infty} frac{(x^2-2x-4)-(x^2-7x+5)}{sqrt{x^2-2x-4}+sqrt{x^2-7x+5}} = limlimits _{x to infty } frac{5x-9}{sqrt{x^2-2x-4}+sqrt{x^2-7x+5}} =\\= limlimits _{x to infty}; frac{xcdot (5-frac{9}{x})}{xcdot left (sqrt{1-frac{2}{x}-frac{4}{x^2}}+sqrt{1-frac{7}{x}+frac{5}{x^2} }right )} =frac{5}{1}=5$