\(=lim\left[\sqrt{n}\left(1-\frac{1}{\sqrt[6]{n}}+\frac{1}{\sqrt{n}}\right)\right]=\infty.1=+\infty\)
\(=lim\left[\sqrt{n}\left(1-\frac{1}{\sqrt[6]{n}}+\frac{1}{\sqrt{n}}\right)\right]=\infty.1=+\infty\)
\(1.lim\left(\sqrt[3]{8n^3+4n^2+1}-\sqrt[3]{8n^3-2}\right)\)
\(2.lim\left(\sqrt[3]{n^3+n^2+1}+\sqrt[3]{8-n^3}\right)\)
\(3.lim\left(\sqrt[3]{n^3+n^2+2}-n\right)\)
1
a,Lim\(\sqrt{1+2n-n^3}\)
b,Lim\(\sqrt{n^2+2n+3}-\sqrt[3]{n^2+n^3}\)
c,Lim\(\dfrac{\left(2\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n+1\right)\left(n+2\right)}\)
d,\(\dfrac{4^{n+1}-3\times2^n}{3^{n+2}+2^n}\)
e,\(\dfrac{7^{n+1}-5^{n+2}+3}{2\times6^{n+1}-3^n+3}\)
f,\(\dfrac{\sqrt{n^4+1}}{n}\) -\(\dfrac{\sqrt{4n^6+1}}{n}\)
tính
a. \(\lim\limits_{n->+\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
b.\(\lim\limits_{n->+\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
lim \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\)
lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
\(\lim\limits\left[\left(1-n\right)\left(\sqrt{n^2-6n}-\sqrt[3]{n^3-27n^2}\right)\right]\)
tính
a.\(\lim\limits_{n->+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)
b.\(\lim\limits_{n->+\infty}\dfrac{\sqrt{n^2-n+2}}{n+2}\)
c.\(\lim\limits_{n->+\infty}\dfrac{n-\sqrt[3]{n^2-n^3}}{n^2+n+1}\)
d.\(\lim\limits_{n->+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
lim(x\(\rightarrow\)1)\(\dfrac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right).....\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{n-1}}\)
a)lim \(\frac{\left(2n+1\right)^2\left(n-1\right)}{\sqrt[3]{n^3+7n-2}}\)
b)lim [(2n-1)\(\sqrt{\frac{2n^2+5}{n^4+n^2+2}}\)]
c)lim [n(\(\sqrt[3]{n^3+n^2}-n\))]
tính các giới hạn sau:
a, lim\(\frac{n^{2020}-n+1}{n^{2022}+2n-3}\)
b, lim(\(\sqrt[3]{n^3-2n^2}-n\))
c, lim \(\left(\sqrt{n^2+3n}-n+2\right)\)
d, lim \(n\left(\sqrt{n^2-1}-\sqrt{n^2+2}\right)\)
Tính: Lim\(\left(\sqrt{n^2+2}.\sqrt[3]{8n^3+1}-\sqrt{4n^2+1}.\sqrt[3]{n^3+2}\right)\)