5.1. Find the limit of the sequence

$$
\lim _{n \rightarrow \infty} \frac{3 n^{5}-6 n^{4}+3 n+5}{(n+1)^{6}-(n-2)^{6}}
$$
Solution. First, we transform the denominator of this sequence using the formulas for abbreviated multiplication:

$$
\begin{gathered}
(n+1)^{6}-(n-2)^{6}=\left((n+1)^{3}-(n-2)^{3}\right)\left((n+1)^{3}+(n-2)^{3}\right)= \\
=3\left((n+1)^{2}+(n+1)(n-2)+(n-2)^{2}\right)\left((n+1)^{2}-(n+1) \times\right. \\
\left.\times(n-2)+(n-2)^{2}\right)(2 n-1)= \\
=3(2 n-1)\left(3 n^{2}-3 n+3\right)\left(n^{2}-n+7\right)
\end{gathered}
$$

Substitute this result into the original limit and divide the numerator and the denominator of the fraction by $n^{5}$. After this, discarding the infinitesimals, we obtain the required limit:

$$
\lim _{n \rightarrow \infty} \frac{3-6 / n+3 / n^{4}+5 / n^{5}}{3(2-1 / n)\left(3-3 / n+3 / n^{2}\right)\left(1-1 / n+7 / n^{2}\right)}=1 / 6
$$