\(\lim\limits\left(\sqrt[3]{1+2n-n^3}-n\right)\)
\(=\lim\limits\dfrac{1+2n-n^3-n^3}{\sqrt[3]{\left(1+2n-n^3\right)^2}+n\cdot\sqrt[3]{1+2n-n^3}+n^2}\)
\(=\lim\limits\dfrac{1+2n-2n^3}{\sqrt[3]{\left(1+2n-n^3\right)^2}+n\cdot\sqrt[3]{1+2n-n^3}+n^2}\)
\(=\lim\limits\dfrac{n^3\left(-2+\dfrac{2}{n^2}+\dfrac{1}{n^3}\right)}{n^2\cdot\sqrt[3]{\left(\dfrac{1}{n^3}+\dfrac{2}{n^2}-1\right)^2}+n^2\cdot\sqrt[3]{-1+\dfrac{2}{n^2}+\dfrac{1}{n^3}}+n^2}\)
\(=\lim\limits\dfrac{n\left(-2+\dfrac{2}{n^2}+\dfrac{1}{n^3}\right)}{\sqrt[3]{\left(\dfrac{1}{n^3}+\dfrac{2}{n^2}-1\right)^2}+\sqrt[3]{-1+\dfrac{2}{n^2}+\dfrac{1}{n^3}}+1}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits n=+\infty\\\lim\limits\dfrac{-2+\dfrac{2}{n^2}+\dfrac{1}{n^3}}{\sqrt[3]{\left(\dfrac{1}{n^3}+\dfrac{2}{n^2}-1\right)^2}+\sqrt[3]{-1+\dfrac{2}{n^2}+\dfrac{1}{n^3}}+1}=\dfrac{-2}{1+1+1}=-\dfrac{2}{3}< 0\end{matrix}\right.\)
Phương pháp: Nhân cả tử và mẫu với biểu thức liên hợp
1. \(A\pm B\) có liên hợp là \(A\mp B\).
2. \(\sqrt{A}\pm B\) có liên hợp là \(\sqrt{A}\mp B\).
3. \(\sqrt{A}\pm\sqrt{B}\) có liên hợp là \(\sqrt{A}\mp\sqrt{B}\).
4. \(\sqrt[3]{A}\pm B\) có liên hợp là \(\sqrt[3]{A^2}\mp B\sqrt[3]{A}+B^2\).
Bài giải: Áp dụng biểu thức liên hợp số 4
\(\lim\left(\sqrt[3]{1+2n-n^3}-n\right)\)
\(=\lim\dfrac{\left(\sqrt[3]{1+2n-n^3}-n\right)\left[\sqrt[3]{\left(1+2n-n^3\right)^2}+n\sqrt[3]{1+2n-n^3}+n^2\right]}{\sqrt[3]{\left(1+2n-n^3\right)^2}+n\sqrt[3]{1+2n-n^3}+n^2}\)
\(=\lim\dfrac{1+2n-n^3-n^3}{\sqrt[3]{n^6-4n^4-2n^3+4n^2+4n+1}+\sqrt[3]{n^3+2n^4-n^6}+n^2}\)
\(=\lim\dfrac{\left(1+2n-2n^3\right)\div n^3}{\left(\sqrt[3]{n^6-4n^4-2n^3+4n^2+4n+1}+\sqrt[3]{n^3+2n^4-n^6}+n^2\right)\div n^3}\)
\(=\lim\dfrac{\dfrac{1}{n^3}+\dfrac{2}{n^2}-2}{\sqrt[3]{\dfrac{1}{n^3}-\dfrac{4}{n^5}-\dfrac{2}{n^6}+\dfrac{4}{n^7}+\dfrac{4}{n^8}+\dfrac{1}{n^9}}+\sqrt[3]{\dfrac{1}{n^6}+\dfrac{2}{n^5}-\dfrac{1}{n^3}}+\dfrac{1}{n}}\)
\(=-\infty\)
(vì \(\lim\left(\dfrac{1}{n^3}+\dfrac{2}{n^2}-2\right)=-2\) và \(\lim\left(\sqrt[3]{\dfrac{1}{n^3}-\dfrac{4}{n^5}-\dfrac{2}{n^6}+\dfrac{4}{n^7}+\dfrac{4}{n^8}+\dfrac{1}{n^9}}+\sqrt[3]{\dfrac{1}{n^6}+\dfrac{2}{n^5}-\dfrac{1}{n^3}}+\dfrac{1}{n}\right)=0\), chia được \(\dfrac{-2}{0}\) nên ra \(-\infty\))
