\(lim\frac{\sqrt{n^2+1}+\sqrt{n+1}}{3n+2}=lim\frac{\sqrt{1+\frac{1}{n^2}}+\sqrt{\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{2}{n}}=\frac{1}{3}\)
\(lim\frac{\sqrt{n^2+1}+\sqrt{n+1}}{3n+2}=lim\frac{\sqrt{1+\frac{1}{n^2}}+\sqrt{\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{2}{n}}=\frac{1}{3}\)
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 \(\dfrac{\sqrt{n+1}}{1+\sqrt{n}}\)
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 \(\dfrac{(2)^{n}+(3)^{n+2}}{4×(3)^{n}+(2)^{n+3}}\)
Tìm \(\lim\limits_{x->-\infty}\)\(\frac{\left|x\right|\sqrt{4x^2+3}}{2x-1}\)
lim \(\sqrt{n}\)(\(\sqrt{n+4}\)-\(\sqrt{n+3}\))
lim (n-2-\(\sqrt{3n^2+n-1}\))
\(\lim\limits_{x->0}\)\(\frac{\sqrt[3]{x^3-2x+1}-1}{x^2+2x}\)
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 \(n\left(\sqrt{n^2-1}-\sqrt{n^2+2}\right)\)
a; lim\(\frac{\sqrt{6n^4+n+1}}{2n^2+1}\)
b; lim \(\frac{\left(n+1\right)\left(2n+1\right)^2\left(3n+1\right)^3}{n^2\left(n+2\right)^2\left(1-3n\right)^2}\)
Tìm giới hạn lim un
a. \(u_n=\left(2-3n\right)^4\left(n+1\right)^3\)
b.\(u_n=\sqrt[3]{n+4}-\sqrt[3]{n+1}\)
c.\(u_n=\sqrt[3]{8n^3+3n^2+4}-2n+6\)
d. \(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\)
Help me ! Gợi ý cho mik cx đc ạ . Tks mng
lim n(\(\sqrt[3]{n^3-3n^2}-3n\))
lim (\(\sqrt{4n^2+n}+\sqrt[3]{2n^2-8n^3}\))
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.\(lim\dfrac{3^n-4.2^{n-1}-10}{7.2^n+4^n}\)
\(lim\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+......+\frac{1}{\sqrt{n^2+n}}\right)\)