Zadacha Kuznecov Predely 2-4

Условие задачи

Вычислить предел числовой последовательности:

$\lim _{n\to \infty }{\frac {(1-n)^{4}-(1+n)^{4}}{(1+n)^{3}-(1-n)^{3}}}$

Решение

$\lim _{n\to \infty }{\frac {(1-n)^{4}-(1+n)^{4}}{(1+n)^{3}-(1-n)^{3}}}=$

$=\lim _{n\to \infty }{\frac {\left((1-n)^{2}-(1+n)^{2}\right)\cdot \left((1-n)^{2}+(1+n)^{2}\right)}{1+3n+3n^{2}+n^{3}-1+3n-3n^{2}+n^{3}}}=$

$=\lim _{n\to \infty }{\frac {(1-2n+n^{2}-1-2n-n^{2})\cdot (1-2n+n^{2}+1+2n+n^{2})}{6n+2n^{3}}}=$

$=\lim _{n\to \infty }{\frac {-4n(2+2n^{2})}{2n(3+n^{2})}}=$

$=\lim _{n\to \infty }{\frac {-4(1+n^{2})}{3+n^{2}}}=\lim _{n\to \infty }{\frac {n^{2}\left(-4\left({\frac {1}{n^{2}}}+1\right)\right)}{n^{2}\left({\frac {3}{n^{2}}}+1\right)}}=$

$=\lim _{n\to \infty }{\frac {-4\left({\frac {1}{n^{2}}}+1\right)}{{\frac {3}{n^{2}}}+1}}={\frac {-4(0+1)}{0+1}}=-4$