Срочно 20 баллов, 1-ую плиз!. Created by jengo97. algebra-ru

1) 1 способ:2 способ. Можно по правилу Лопиталя:2) 1 способ:2 способ. Правило Лопиталя:

Срочно 20 баллов, 1-ую плиз!...

jengo97 @jengo97

July 2022 1 6 Report

Срочно 20 баллов, 1-ую плиз!

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kalbim

Verified answer

1) 1 способ:
$\lim_{n \to \infty} \frac{7n^{2}+6n+5}{n^{3}-5n+5} = \lim_{n \to \infty} \frac{ \frac{7}{n}+ \frac{6}{n^{2}}+ \frac{5}{n^{3}}}{1- \frac{5}{n^{2}}+ \frac{5}{n^{3}}}= \lim_{n \to \infty} \frac{ \frac{7}{n}+ \frac{6}{n^{2}}+ \frac{5}{n^{3}}}{1- \frac{5}{n^{2}}+ \frac{5}{n^{3}}}=0$

2 способ. Можно по правилу Лопиталя:
$\lim_{n \to \infty} \frac{7n^{2}+6n+5}{n^{3}-5n+5} =\lim_{n \to \infty} \frac{14n+6}{3n^{2}-5} =\lim_{n \to \infty} \frac{14}{6n}=0$

2) 1 способ:
$\lim_{n \to \infty} \frac{7n^{3}+6n^{2}+5}{n^{4}-5n^{2}+5} =lim_{n \to \infty} \frac{ \frac{7}{n}+ \frac{6}{n^{2}}+ \frac{5}{n^{4}}}{1- \frac{5}{n^{2}}+ \frac{5}{n^{4}}}=0$

2 способ. Правило Лопиталя:
$\lim_{n \to \infty} \frac{7n^{3}+6n^{2}+5}{n^{4}-5n^{2}+5} =\lim_{n \to \infty} \frac{21n^{2}+12n}{4n^{3}-10n}=\lim_{n \to \infty} \frac{42n+12}{12n^{2}-10}=\lim_{n \to \infty} \frac{42}{24n}$

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