Cách làm cho bạn:
a) \(\lim({n^3} + 2{n^2}-n + 1)= \lim n^3(1 + \frac{2}{n}-\frac{1}{n^{2}}+\frac{1}{n^{3}}) = +\infty \)
b) \(\lim( - {n^2} + 5n-2) = \lim -n^2 ( 1 - \frac{5}{n}+\frac{2}{n^{2}}) = -\infty \)
c) \(\lim (\sqrt{n^{2}-n} - n) = \lim \frac{(\sqrt{n^{2}-n}-n)(\sqrt{n^{2}-n}+n)}{\sqrt{n^{2}-n}+n}\)
\(= \lim \frac{n^{2}-n-n^{2}}{\sqrt{n^{2}-n}+n}\)
\(= \lim \frac{-n}{\sqrt{{n^2}\left( {1 - {1 \over n}} \right)}+ n}\)
\(= \lim \frac{-1}{\sqrt{1-\frac{1}{n}}+1}\)
\(=\frac{-1}{1+1}= \frac{-1}{2}\).
d) \(\lim (\sqrt{n^{2}-n} + n) = \lim \left( {\sqrt {{n^2}\left( {1 - {1 \over n}} \right)} + n} \right) \)
\(= \lim n.\left( {\sqrt {1 - {1 \over n}} + 1} \right)= +\infty \).
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