\(lim\frac{1}{\sqrt{n^2+1}-\sqrt{n+2}}=lim\frac{\frac{1}{n}}{\sqrt{1+\frac{1}{n^2}}-\sqrt{\frac{1}{n}+\frac{2}{n^2}}}=\frac{0}{1-0}=0\)
\(lim\left(\sqrt{n^2+2n+2}+n\right)=lim\left[n\left(\sqrt{1+\frac{2}{n}+\frac{2}{n^2}}+1\right)\right]=\infty\left(1+1\right)=+\infty\)
a, lim \(\dfrac{\sqrt{n+1}}{1+\sqrt{n}}\)
b, lim \(\dfrac{1+2+...+n}{n^2+2}\)
c, lim \((\sqrt{n^2+n+1}-n)\)
d, lim \((\sqrt{3n-1}-\sqrt{2n-1})\)
e, lim \((\sqrt[3]{n^3+2n^2}-n)\)
g, lim \(\dfrac{(2)^{n}+(3)^{n+2}}{4×(3)^{n}+(2)^{n+3}}\)
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 \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
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(\(\sqrt[3]{n^3+n^2}-n\))]
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 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
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.\(\lim\limits_{n->+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
Tìm a,b để:
1, Lim(\(\sqrt{4n^2+2n+1}\)-an+b)=1
2, Lim( \(\sqrt{n^2+6n-1}-\sqrt{an^2+bn+2}\))=4
Tính lim(\(\sqrt{4n^2+2}.\sqrt[3]{n^3+1}-2n\sqrt[3]{n^3+2}\))
tính giới hạn
1.\(\lim\limits\left(n^3+4n^2-1\right)\)
2.\(lim\dfrac{\left(n+1\right)\sqrt{n^2-n+1}}{3n^2+n}\)
3.\(lim\dfrac{1+2+....+n}{2n^2}\)
4.\(lim\dfrac{3^n-4.2^{n-1}-10}{7.2^n+4^n}\)
