\(=lim\frac{\sqrt{2+\frac{3}{n}}}{\sqrt{2}+\frac{5}{\sqrt{n}}}=\frac{\sqrt{2+0}}{\sqrt{2}+0}=1\)
Đúng 0
Bình luận (0)
Chương 4: GIỚI HẠN
Các câu hỏi tương tự
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(\(\sqrt[3]{n^3+n^2}-n\))]
\(\lim\limits\frac{\sqrt{n}+\sqrt[3]{n}+\sqrt[4]{n}}{\sqrt{2n+1}}\)
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 \(\dfrac{(2)^{n}+(3)^{n+2}}{4×(3)^{n}+(2)^{n+3}}\)
Tính: lim\(\left(2n-\sqrt{9n^2+n}+\sqrt{n^2+2n}\right)\)
Tính lim(\(\sqrt{4n^2+2}.\sqrt[3]{n^3+1}-2n\sqrt[3]{n^3+2}\))
1 a,Limsqrt{1+2n-n^3}b,Limsqrt{n^2+2n+3}-sqrt[3]{n^2+n^3}c,Limdfrac{left(2sqrt{n}+1right)left(sqrt{n}+3right)}{left(n+1right)left(n+2right)}d,dfrac{4^{n+1}-3times2^n}{3^{n+2}+2^n}e,dfrac{7^{n+1}-5^{n+2}+3}{2times6^{n+1}-3^n+3}f,dfrac{sqrt{n^4+1}}{n} -dfrac{sqrt{4n^6+1}}{n}
Đọc tiếp
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 \(n\left(\sqrt{n^2-1}-\sqrt{n^2+2}\right)\)
lim \(\frac{\sqrt{n^3+2n}+1}{n+2}\)