\(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\lim\limits\frac{\frac{1}{\sqrt{n}}}{\sqrt{1+\frac{1}{n}}+1}=\frac{0}{2}=0\)
Đúng 0
Bình luận (0)
Chương 4: GIỚI HẠN
Các câu hỏi tương tự
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 \(\dfrac{(2)^{n}+(3)^{n+2}}{4×(3)^{n}+(2)^{n+3}}\)
\(1.lim\left(\sqrt[3]{8n^3+4n^2+1}-\sqrt[3]{8n^3-2}\right)\)
\(2.lim\left(\sqrt[3]{n^3+n^2+1}+\sqrt[3]{8-n^3}\right)\)
\(3.lim\left(\sqrt[3]{n^3+n^2+2}-n\right)\)
1 a,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
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 \(\frac{1}{\sqrt{n^2+1}-\sqrt{n+2}}\)
lim \(\sqrt{n^2+2n+2}+n\)
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-\(\sqrt{3n^2+n-1}\))
\(\lim\limits_{x->0}\)\(\frac{\sqrt[3]{x^3-2x+1}-1}{x^2+2x}\)
\(lim=\frac{\sqrt{n+\sqrt{n+1}}}{\sqrt{n-\sqrt{n}}}\)
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 \(n\left(\sqrt{n^2-1}-\sqrt{n^2+2}\right)\)
tínha.limlimits_{n-+infty}dfrac{n^5+n^2-n+2}{left(2n^3-1right)left(n^2+n+1right)}b.limlimits_{n-+infty}dfrac{sqrt{n^2-n+2}}{n+2}c.limlimits_{n-+infty}dfrac{n-sqrt[3]{n^2-n^3}}{n^2+n+1}d.limlimits_{n-+infty}left(n-sqrt{n^2+n+1}right)
Đọc tiếp
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.\(\lim\limits_{n->+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
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(\(\sqrt[3]{n^3+n^2}-n\))]