Tìm các giới hạn sau:
a/ $lim(\sqrt[3]{2n-n^{3}}+ n -1)$
b/ $lim\frac{n^{2}+\sqrt[3]{1-n^{6}}}{\sqrt{n^{4}+1}-n^{2}}$
2/ Rút gọn: $\frac{1}{1\sqrt{2}+2\sqrt{1}}+ \frac{1}{2\sqrt{3}+3\sqrt{2}}+ ... + \frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}$
Tìm lim $u_{n}$
'Hầu hết' đều dùng kỹ thuật 'liên hiệp'.
a) $\sqrt[3]{2n-n^{3}}+ n -1 =\frac{ (2n-n^{3})+ (n -1)^3}{\sqrt[3]{(2n-n^{3})^2}-(n-1)\sqrt[3]{2n-n^{3}}+(n-1)^2}=\frac{- 3n^2 + 5n - 1}{\sqrt[3]{(2n-n^{3})^2}-(n-1)\sqrt[3]{2n-n^{3}}+(n-1)^2}.$
Do đó
$\sqrt[3]{2n-n^{3}}+ n -1 =\frac{- 3 + 5/n - 1/n^2}{\sqrt[3]{(2/n^2-1)^2}-(1-1/n)\sqrt[3]{2/n^2-1}+(1-1/n)^2}.$
Suy ra $\lim \left( \sqrt[3]{2n-n^{3}}+ n -1 \right)= \frac{-3}{1+1+1}=-1.$
b) Làm tương tự cho tử, mẫu.
c) Nhận xét:
$\frac{1}{k\sqrt{k+1}+(k+1)\sqrt{k}}= \frac{1}{\sqrt{k}\sqrt{k+1}\left(\sqrt{k+1}+\sqrt{k}\right)}= \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1} }= \frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\, \forall k>0.$