Question

!!!!!!!!!!!!!!Помогите с решением лимитов( с объяснением)7,9,10,12 ПРИМЕРЫ

Answer 1

$7)\; \; \lim\limits _{x \to 8}\frac{\sqrt{9+2x}-5}{\sqrt[3]{x}-2}=\lim\limits _{x \to 8}\frac{(9+2x-25)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{(x-8)(\sqrt{9+2x}+5)}=\\\\=\lim\limits _{x \to 8}\frac{2(x-8)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{(x-8)(\sqrt{9+2x}+5)}=\frac{2(\sqrt[3]{8^2}+2\sqrt[3]{8}+4)}{\sqrt{25}+5}=\frac{2(4+4+4)}{10}=\frac{12}{5}$

$9)\; \; \lim\limits _{x \to 0}\frac{7x}{sinx+sin7x}=\lim\limits _{x \to 0}\frac{7x}{2sin4x\cdot cos3x}=\Big [\, sin\alpha \sim \alpha \; ,\; \alpha \to 0\; \Big ]=\\\\=\lim\limits _{x \to 0}\frac{7x}{2\cdot 4x\cdot cos3x}=\frac{7}{2\cdot 4\cdot 1}=\frac{7}{8}\\\\10)\; \; \lim\limits _{x \to 0}\frac{cosx-cos^3x}{5x^2}=\lim\limits _{ x\to 0}\frac{cosx\cdot (1-cos^2x}{5x^2}=\lim\limits _{x \to 0}\frac{-cosx\cdot sin^2x}{5x^2}=\\\\=\lim\limits _{x \to 0}\frac{cosx\cdot x^2}{5x^2}=\lim\limits _{x \to 0}\frac{cosx}{5}=\frac{1}{5}$

$12)\; \; \lim\limits _{x \to \infty }\Big (\frac{x}{x-1}\Big )^{3-2x}=\lim\limits _{x \to \infty }\Big (1+\frac{1}{x-1}\Big )^{\frac{x-1}{1}\cdot \frac{3-2x}{x-1}}=\\\\=\lim\limits _{x \to \infty }\Big (\Big (1+\frac{1}{x-1}\Big )^{x-1}\Big )^{\frac{3-2x}{x-1}}=e^{\lim\limits _{x \to \infty}\frac{3-2x}{x-1}}=e^{-2}=\frac{1}{e^2}$