Question

Решите замечательный предел

Answer 1

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

$2); ; limlimits_{x to infty} (sqrt{2x^2-3}-5x)= limlimits _{x to infty} frac{ sqrt{2x^2-3}+5x }{2x^2-3-25x^2} = limlimits _{x to infty}frac{sqrt{2x^2-3}+5x}{-3-23x^2}=\\= limlimits _{x to infty}frac{sqrt{frac{2}{x^4}-frac{3}{x^4}}+frac{5}{x}}{-frac{3}{x^2}-23} = frac{0}{-23} =0$

$3); ; limlimits _{x to 0}frac{2x, sinx}{secx-1} =limlimits _{x to 0} frac{2x, sinx}{frac{1}{cosx}-1} = limlimits _{x to 0}frac{2x, sinxcdot cosx}{1-cosx} =\\= limlimits_{x to 0}frac{xcdot sin2x}{2sin^2frac{x}{2}} = limlimits _{x to 0}frac{xcdot 2x}{2cdot (frac{x}{2})^2} = limlimits _{x to o} frac{2x^2cdot 4}{2x^2} =4$

Answer 2

Большое тебе спасибо