Nội dung bài giảng
Bài 7. Tính các giới hạn sau:
a) \(\lim({n^3} + {\rm{ }}2{n^2}-{\rm{ }}n{\rm{ }} + {\rm{ }}1)\);
b) \(\lim{\rm{ }}( - {n^2} + {\rm{ }}5n{\rm{ }}-{\rm{ }}2)\);
c) \(\lim (\sqrt{n^{2}-n}- n)\);
d) \(\lim (\sqrt{n^{2}-n} + n)\).
Giải:
a) \(\lim({n^3} + {\rm{ }}2{n^2}-{\rm{ }}n{\rm{ }} + {\rm{ }}1)= \lim n^3(1 + \frac{2}{n}-\frac{1}{n^{2}}+\frac{1}{n^{3}}) = +∞\)
b) \(\lim{\rm{ }}( - {n^2} + {\rm{ }}5n{\rm{ }}-{\rm{ }}2) = \lim n^2 ( -1 + \frac{5}{n}-\frac{2}{n^{2}}) = -∞\)
c) \(\lim (\sqrt{n^{2}-n} - n) = \lim \frac{(\sqrt{n^{2}-n}-n)(\sqrt{n^{2}-n}+n)}{\sqrt{n^{2}-n}+n}\)
\(= \lim \frac{n^{2}-n-n^{2}}{\sqrt{n^{2}-n}+n} = \lim \frac{-n}{\sqrt{{n^2}\left( {1 - {1 \over n}} \right)}+ n} = \lim \frac{-1}{\sqrt{1-\frac{1}{n}}+1} = \frac{-1}{2}\).
d) \(\lim (\sqrt{n^{2}-n} + n) = \lim \left( {\sqrt {{n^2}\left( {1 - {1 \over n}} \right)} + n} \right) \)
\(= \lim n.\left( {\sqrt {1 - {1 \over n}} + 1} \right)= +∞\).