Bài 7. Tính các giới hạn sau:
a) lim;
b) \lim{\rm{ }}( - {n^2} + {\rm{ }}5n{\rm{ }}-{\rm{ }}2);
c) \lim (\sqrt{n^{2}-n}- n);
d) \lim (\sqrt{n^{2}-n} + n).
Advertisements (Quảng cáo)
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)= +∞.