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March 2022 1 34 Report

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NNNLLL54

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$1)\; \lim\limits _{x \to -2}\frac{x^2-2x-8}{x^3+8}=\lim\limits_{x \to -2}\frac{(x+2)(x-4)}{(x+2)(x^2-2x+4)}=\lim\limits_{x \to -2}\frac{x-4}{x^2-2x+4}=\\=\frac{-2-4}{4+4+4}= \frac{-6}{12}=-\frac{1}{2}\\\\2)\; \lim\limits_{x \to \infty}\Big (\frac{7x-4}{7x+1}\Big )^{2x-5}=\lim\limits_{x \to \infty}\Big (\Big (1+\frac{-5}{7x+1}\Big )^{\frac{7x+1}{-5}}\Big )^{\frac{-5(2x-5)}{7x+1}}=\\\\=e^{\lim\limits_{x \to \infty}\frac{-10x+25}{7x+1}}=e^{-\frac{10}{7}}$

$3)\; \lim\limits_{x \to 1} \frac{2-\sqrt{x}}{\sqrt{6x+1}-5}=\frac{2-\sqrt1}{\sqrt{7}-5}=\frac{1}{\sqrt7-5}\\\\4)\; \lim\limits_{x \to \infty }\frac{8x^2-9x+5}{4x^2-6x+7}= \lim\limits_{x \to \infty}\frac{8x^2}{4x^2}=\frac{8}{4}=2\\\\5)\; \lim\limits_{x \to \infty}(\sqrt{x^2-5x+7}-x)=\lim\limits_{x \to \infty}\frac{(\sqrt{x^2-5x+7}-x)(\sqrt{x^2-5x+7}+x)}{\sqrt{x^2-5x+7}+x}=\\\\=\lim\limits_{x \to \infty}\frac{x^2-5x+7-x^2}{\sqrt{x^2-5x+7}+x}=\lim\limits _{x \to \infty}\frac{7-5x}{\sqrt{x^2-5x+7}+x}}=$

$= \lim\limits_{x \to \infty}\frac{\frac{7}{x}-5}{\sqrt{1-\frac{5}{x}+\frac{7}{x^2}}+1}=\frac{-5}{1+1}=-2,5\\\\6)\; \lim\limits _{x \to 0}\frac{arcsin7x}{5x}=[\, arcsin \alpha (x)\sim \alpha (x),\; \alpha (x)\to 0\, ]=\\\\=\lim\limits _{x \to 0}\, \frac{7x}{5x}=\frac{7}{5}$

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