\(=\lim\dfrac{1+2+3+...+n}{2\left(n^2+1\right)}=\lim\dfrac{n\left(n+1\right)}{4\left(n^2+1\right)}=\lim\dfrac{1+\dfrac{1}{n}}{4+\dfrac{4}{n^2}}=\dfrac{1}{4}\)
\(=\lim\dfrac{1+2+3+...+n}{2\left(n^2+1\right)}=\lim\dfrac{n\left(n+1\right)}{4\left(n^2+1\right)}=\lim\dfrac{1+\dfrac{1}{n}}{4+\dfrac{4}{n^2}}=\dfrac{1}{4}\)
giá trị của giới hạn lim \(\dfrac{\dfrac{1}{2}+1+\dfrac{3}{2}+...+\dfrac{n}{2}}{n^2+1}\)
Tìm các giới hạn sau:
a) \(lim\sqrt[3]{-n^3+2n^2-5}\)
b) \(lim\dfrac{1}{\sqrt{n+1}-\sqrt{n}}\)
c) \(lim\left(\dfrac{1}{n+1}-n\right)\)
d) \(lim\left(\dfrac{2n^2-1}{n+1}-2n\right)\)
e) \(lim\dfrac{2n^3+n^2-3n+1}{2-3n}\)
Tìm \(lim\) \(u_n\), biết \(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\).
A. \(lim\) \(u_n=\dfrac{3}{4}\).
B. \(lim\) \(u_n=\dfrac{3}{5}\).
C. \(lim\) \(u_n=\dfrac{2}{3}\).
D. \(lim\) \(u_n=\dfrac{4}{3}\).
Giải thích chi tiết bước làm và tại sao lại làm như vậy.
Tìm giới hạn dãy số :
\(a,lim\dfrac{5n+1}{2n}\\ b,lim\dfrac{6n^2+8n+1}{5n^2+3}\\ c,lim\dfrac{3^n+2^n}{4.3^n}\\ d,lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}\)
lim \(\dfrac{\dfrac{1}{2}+1+\dfrac{3}{2}+.....+\dfrac{n}{2}}{n^2+1}\)
Tìm các giới hạn sau:
a)\(lim\left[n^2\left(\sqrt{n^2+2}-\sqrt{n^2+4}\right)\right]\)
b)lim( \(\dfrac{3}{n-2}-5n\))
c) lim(\(\dfrac{n-1}{\sqrt{3}-n}-\dfrac{4}{2^{-n}}\))
d) \(lim\left(\dfrac{n^2-4}{n-2}-\dfrac{3n^2+4}{n}\right)\)
e) \(lim\dfrac{\sqrt{n^2+1}-n\sqrt{5}}{\sqrt{n^2+1}+n\sqrt{5}}\)
Tính:
A= \(lim\dfrac{n+1}{n^2+2n}\)
B= \(lim\left(-2n^3+n^2+2\right)\)
C= \(lim\dfrac{\sqrt{9n^2-n-1}}{4n-2}\)
D= \(lim\dfrac{3^n+5.4^n}{4^n+2^n}\)
Tính các giới hạn sau
1,Lim\(\left(\dfrac{2n^3}{2n^2+3}+\dfrac{1-5n^2}{5n+1}\right)\)
2,a,Lim\(\left(\sqrt{n^2+n}-\sqrt{n^2+2}\right)\)
b,Lim\(\dfrac{\sqrt{n^4+3n-2}}{2n^2-n+3}\)
c,Lim\(\dfrac{\sqrt{n^2-4n}-\sqrt{4n^2+1}}{\sqrt{3n^2+1}-n}\)
Tính \(lim\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{2}{5}+\left(\dfrac{2}{5}\right)^2+...+\left(\dfrac{2}{5}\right)^n}\)