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Округлите до сотых: 197,203 359,855 63,562 12,396

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Pustynya
4,7(21 оценок)

197,203=197,20 359,855=359,86 63,562=63,56 12,396=12,40

Пошаговое объяснение:

\displaystyle 1)\, \lim_{x\to\infty}\frac{5n+n^2}{3-7n}=\lim_{x\to\infty}\frac{(5n+n^2)}{(3-7n)}=\lim_{x\to\infty}\frac{n^-^2(5n+n^2)}{n^-^2(3-7n)}=\lim_{x\to\infty}\frac{\frac1{n^2}(5n+n^2)}{\frac1{n^2}(3-7n)}=-\lim_{x\to\infty}\frac{\frac1{n^2}*5n+\frac1{n^2}*n^2}{\frac1{n^2}*3-\frac1{n^2}*7n}=-\lim_{x\to\infty}\frac{\frac{5n}{n^2}+\frac{n^2}{n^2}}{\frac3{n^2}-\frac{7n}{n^2}}=-\lim_{x\to\infty}\frac{\frac{5}{n}+1}{\frac3{n^2}-\frac{7}{n}}=-\frac{\frac{5}{\infty}+1}{\frac3{\infty^2}-\frac{7}{\infty}}=

\displaystyle- \frac{0+1}{0-0}=-\frac{1}0=-\infty

По определению: \displaystyle\forall\varepsilon0:\exists N(\varepsilon)\in\mathbb{N}:\forall n\geq N\Rightarrow \frac{5n+n^2}{3-7n}0:\frac{5n+n^2}{3-7n}\varepsilon\Leftrightarrow5n+n^23\varepsilon-7n\varepsilon\\\beth N=\frac{-5+7\varepsilon+\sqrt{(5+7\varepsilon)^2+12\varepsilon}}{2}, \because\varepsilon0\, \wedge\, -577\varepsilon\Rightarrow (5+7\varepsilon)^2+12\varepsilon0.\because\forall n\geq N,\frac{5n+n^2}{3-7n}

ЧТД

\displaystyle 2)\, \lim_{n \to \infty}\frac{3-2\sqrt n}{1-5\sqrt n}=\lim_{n \to \infty}\frac{\frac{1}{\sqrt n}(3-2\sqrt n)}{\frac{1}{\sqrt n}(1-5\sqrt n)}=\lim_{n \to \infty}\frac{\frac{3}{\sqrt n}-\frac{2\sqrt n}{\sqrt n}}{\frac{1}{\sqrt n}-\frac{5\sqrt n}{\sqrt n}}=\lim_{n \to \infty}\frac{\frac{3}{\sqrt n}-2}{\frac{1}{\sqrt n}-5}=\frac{\frac3{\infty}-2}{\frac1{\infty}-5}=\frac{0-2}{0-5}=\frac{-2}{-5}=\frac25

По определению:

\displaystyle\forall\varepsilon0:\exists N(\varepsilon)\in\mathbb{N}:\forall n\geq N\Rightarrow \left |\frac{3-2\sqrt n}{1-5\sqrt n}-\frac25\right|0: \left |\frac{3-2\sqrt n}{1-5\sqrt n}-\frac25\right|

\displaystyle-\varepsilon

\displaystyle \beth N=\left | {{t=\frac{\varepsilon-1}{-8-5\varepsilon}} \atop {t=\frac{-\varepsilon-1}{-8-5\varepsilon}}} \right.:\\\because\forall n 0 : \sqrt{n} 0, \forall n\geq N, \left |\frac{3-2\sqrt n}{1-5\sqrt n}-\frac25\right |

ЧТД

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