Tính các giới hạn sau:
a) \(lim (n^3 + 2n^2 - n + 1)\);
b) \(lim (-n^2 + 5n - 2)\);
c) \(lim(\sqrt{n^{2}-n}-n)\) ;
d) (\(lim(\sqrt{n^{2}-n}+n)\).
Câu a:
\(lim (n^3 + 2n^2 - n + 1) = lim \ n^3 (1 + \frac{2}{n}-\frac{1}{n^{2}}+\frac{1}{n^{3}})=+\infty\)
Câu b:
\(lim (-n^2 + 5n - 2) = lim \ n^2 ( -1 + \frac{5}{n}-\frac{2}{n^{2}})=-\infty\)
Câu c:
\(lim (\sqrt{n^2-n}+n)= lim \frac{n^2-n-n^2}{\sqrt{n^2-n }+n}= lim\frac{-n}{\sqrt{n^2-n}+n}\)
\(=-lim\frac{1}{\sqrt{1-\frac{1}{n}}+n}=-\frac{1}{2}.\)
Câu d:
\(lim (\sqrt{n^2-n}+n)=lim \ n(\sqrt{1-\frac{1}{n}}+ 1) = +\infty\)
-- Mod Toán 11
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