\(=lim\dfrac{n^4+2n^3-n^4}{\sqrt{n^4+2n^3}+n^2}=\dfrac{2n^3}{\sqrt{n^4+2n^3}+n^2}\)
Bậc tử lớn hơn bậc mẫu và tích hệ số lớn nhất của tử và mẫu >0 nên lim= dương vô cùng
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.\(lim\dfrac{3^n-4.2^{n-1}-10}{7.2^n+4^n}\)
Tìm các giới hạn sau:
a) \(\lim\limits\left(\sqrt{2n^2+3}-\sqrt{n^2+1}\right)\)
b) \(\lim\limits\dfrac{1}{\sqrt{n+1}-\sqrt{n}}\)
\(\lim\limits\left[\left(1-n\right)\left(\sqrt{n^2-6n}-\sqrt[3]{n^3-27n^2}\right)\right]\)
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 \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
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,\(\dfrac{\sqrt{n^4+1}}{n}\) -\(\dfrac{\sqrt{4n^6+1}}{n}\)
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: lim\(\left(2n-\sqrt{9n^2+n}+\sqrt{n^2+2n}\right)\)
\(\lim\limits\frac{\sqrt{n}+\sqrt[3]{n}+\sqrt[4]{n}}{\sqrt{2n+1}}\)
a; lim\(\frac{\sqrt{6n^4+n+1}}{2n^2+1}\)
b; lim \(\frac{\left(n+1\right)\left(2n+1\right)^2\left(3n+1\right)^3}{n^2\left(n+2\right)^2\left(1-3n\right)^2}\)
