Question

ПОМОГИТЕ РЕШИТЬ ЛИМИТЫ ДАЮ 100 БАЛЛОВ

Answer 1

$\lim_{n \to \infty} \frac{\sqrt[5]{2-3n^4+n^5}-\sqrt[3]{2n^3-1} }{\sqrt{n-2+4n^2}-{\sqrt[3]{1-n^3} } } = \\\lim_{n \to \infty} \frac{\sqrt[5]{\frac{2}{n^5}-\frac{3}{n}+1}-\sqrt[3]{2-\frac{1}{n^3}} }{\sqrt{\frac{1}{n}-\frac{2}{n^2}+4}-{\sqrt[3]{\frac{1}{n^3}-1} } }  = \frac{1-\sqrt[3]{2} }{3}$

$\lim_{x \to 1} \frac{\sqrt{19-3x}-\sqrt{9+7x}}{\sqrt[3]{7x+1}-2} = \\ \lim_{x \to 1} \frac{((19-3x)-(9+7x))(\sqrt[3]{(7x+1)^2}+2\sqrt[3]{7x+1}+4)}{(7x+1-8)\\(\sqrt{19-3x}+\sqrt{9+7x})} = \\ \lim_{x \to 1} \frac{-10(x-1)(\sqrt[3]{(7x+1)^2}+2\sqrt[3]{7x+1}+4)}{7(x-1)(\sqrt{19-3x}+\sqrt{9+7x})} = \\ \lim_{x \to 1} \frac{-10(\sqrt[3]{(7x+1)^2}+2\sqrt[3]{7x+1}+4)}{7(\sqrt{19-3x}+\sqrt{9+7x})} = \frac{-10(4+4+4)}{7(4+4)} = \frac{-120}{56} = -\frac{15}{7}$

$\lim_{x \to 1} (x-1)^2\cot \Pi x = \\ \lim_{x \to 1}\frac{(x-1)^2\cos\Pi x}{\sin\Pi x}  = \\ \lim_{x \to 1}\frac{(x-1)^2\cos\Pi x}{\sin\Pi x}  = \\y = x - 1, x = y + 1, y -> 0\\=  \lim_{y \to 0}\frac{(y)^2\cos\Pi(y+1) }{\sin\Pi (y+1)}  = \\\\\lim_{y \to 0}\frac{y^2\cos\Pi(y+1) }{\sin(\Pi y+\Pi)}  = \\\\\lim_{y \to 0}\frac{y^2\cos\Pi(y+1) }{-\sin(\Pi y)}  = \\\\\lim_{y \to 0}-\frac{y\cos\Pi(y+1) }{\frac{\sin(\Pi y)}{\Pi y} * \Pi}  = 0$