\(\begin{array}{l}
a)\,\,\lim \left( {{n^3} + 2{n^2} - n + 1} \right) \\= \lim {n^3}\left( {1 + \frac{2}{n} - \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)\\
\lim {n^3} = + \infty ;\,\,\lim \left( {1 + \frac{2}{n} - \frac{1}{{{n^2}}} + \frac{1}{{{n^3}}}} \right) = 1 > 0\\
\Rightarrow \lim \left( {{n^3} + 2{n^2} - n + 1} \right) = + \infty
\end{array}\)
\(\begin{array}{l}
b)\,\,\lim \left( { - {n^2} + 5n - 2} \right) = \lim {n^2}\left( { - 1 + \frac{5}{n} - \frac{2}{{{n^2}}}} \right)\\
\lim {n^2} = + \infty \\
\lim \left( { - 1 + \frac{5}{n} - \frac{2}{{{n^2}}}} \right) = - 1 < 0\\
\Rightarrow \lim \left( { - {n^2} + 5n - 2} \right) = - \infty
\end{array}\)
c) \(\lim (\sqrt{n^{2}-n} - n) = \lim \frac{(\sqrt{n^{2}-n}-n)(\sqrt{n^{2}-n}+n)}{\sqrt{n^{2}-n}+n}\)
\(= \lim \frac{n^{2}-n-n^{2}}{\sqrt{n^{2}-n}+n} = \lim \frac{-n}{\sqrt{{n^2}\left( {1 - {1 \over n}} \right)}+ n} \) \(= \lim \frac{-1}{\sqrt{1-\frac{1}{n}}+1} = \frac{-1}{2}\).
\(\begin{array}{l}
d)\,\,\lim \left( {\sqrt {{n^2} - n} + n} \right) = \lim n\left( {\sqrt {1 - \frac{1}{n}} + 1} \right)\\
\lim n = + \infty \\
\lim \left( {\sqrt {1 - \frac{1}{n}} + 1} \right) = 2 > 0\\
\Rightarrow \lim \left( {\sqrt {{n^2} - n} + n} \right) = + \infty
\end{array}\)
