Упр.24.34 ГДЗ Мордкович 10-11 класс (Алгебра)
- а) $$y_n=\frac{(2n+1)(3n-4)-6n^2+12n}{n+5}$$;
б) $$y_n=\frac{n^2(2n+5)-2n^3+5n^2-13}{n(n+1)(n-7)+(1-n)}$$;
в) $$y_n=\frac{(1-n)(n^2+1)+n^3}{n^2+2n}$$;
г) $$y_n=\frac{n(7-n^2)+n^3-3n-1}{(n+1)(n+2)+(2n^2+1)}$$.
а) $$y_n=\frac{(2n+1)(3n-4)-6n^2+12n}{n+5}$$
$$\begin{aligned} y_n&=\frac{6n^2-8n+3n-4-6n^2+12n}{n+5} \\ &=\frac{7n-4}{n+5} \end{aligned}$$
$$\lim_{n\to\infty} y_n=\lim_{n\to\infty}\frac{7n-4}{n+5} =\lim_{n\to\infty}\frac{7-\frac{4}{n}}{1+\frac{5}{n}}=7$$
Ответ: $$7$$
б) $$y_n=\frac{n^2(2n+5)-2n^3+5n^2-13}{n(n+1)(n-7)+(1-n)}$$
$$\begin{aligned} y_n&=\frac{2n^3+5n^2-2n^3+5n^2-13}{n^3-7n^2+n^2-7n+1-n} \\ &=\frac{10n^2-13}{n^3-6n^2-8n+1} \end{aligned}$$
$$\lim_{n\to\infty} y_n =\lim_{n\to\infty}\frac{10n^2-13}{n^3-6n^2-8n+1} =\lim_{n\to\infty}\frac{\frac{10}{n}-\frac{13}{n^3}}{1-\frac{6}{n}-\frac{8}{n^2}+\frac{1}{n^3}}=0$$
Ответ: $$0$$
в) $$y_n=\frac{(1-n)(n^2+1)+n^3}{n^2+2n}$$
$$\begin{aligned} y_n&=\frac{n^2+1-n^3-n+n^3}{n^2+2n} \\ &=\frac{n^2-n+1}{n^2+2n} \end{aligned}$$
$$\lim_{n\to\infty} y_n =\lim_{n\to\infty}\frac{n^2-n+1}{n^2+2n} =\lim_{n\to\infty}\frac{1-\frac{1}{n}+\frac{1}{n^2}}{1+\frac{2}{n}}=1$$
Ответ: $$1$$
г) $$y_n=\frac{n(7-n^2)+n^3-3n-1}{(n+1)(n+2)+(2n^2+1)}$$
$$\begin{aligned} y_n&=\frac{7n-n^3+n^3-3n-1}{n^2+3n+2+2n^2+1} \\ &=\frac{4n-1}{3n^2+3n+3} \end{aligned}$$
$$\lim_{n\to\infty} y_n =\lim_{n\to\infty}\frac{4n-1}{3n^2+3n+3} =\lim_{n\to\infty}\frac{\frac{4}{n}-\frac{1}{n^2}}{3+\frac{3}{n}+\frac{3}{n^2}}=0$$
Ответ: $$0$$
