Question

решите только 4 пример

Answer 1

$1)\; \; \lim\limits _{x \to \infty}\frac{3x-5x^2+2}{2x^3+x+3}=\lim\limits _{x \to \infty}\frac{-5x^2}{2x^3}=\lim\limits _{x \to \infty}\frac{-5}{2x}=\Big [\frac{const}{\infty }\Big ]=0\\\\2)\; \; \lim\limits _{x \to 0}\frac{1-cos2x}{x\cdot sinx}=\lim\limits _{x \to 0}\frac{2sin^2x}{x\cdot x}=\lim\limits _{x \to 0}\frac{2x^2}{x^2}=2\\\\3)\; \; \lim\limits_{x \to 3}\frac{9-x^2}{\sqrt{3x}-3}=\lim\limits _{x \to 3}\frac{(3-x)(3+x)}{\sqrt3\cdot (\sqrt{x}-\sqrt3)}=\lim\limits _{x \to 3}\frac{(\sqrt3-\sqrt{x})(\sqrt3+\sqrt{x})(3+x)}{-\sqrt3\cdot (\sqrt3-\sqrt{x})}=$

$=\lim\limits _{x \to 3}\frac{(\sqrt3+\sqrt{x})(3+x)}{-\sqrt3}=\frac{2\sqrt3\cdot 6}{-\sqrt3}=-12\\\\4)\; \; \lim\limits _{x \to 0}\frac{sin7x}{\sqrt{x+1}-1}=\lim\limits _{x \to 0}\frac{7x\cdot (\sqrt{x-1}+1)}{x+1-1}=\lim\limits _{x \to 0}\frac{7\cdot (\sqrt{x+1}+1)}{1}=7\cdot 2=14$