Lời giải:
\(\lim \frac{1}{1^2+1}.\frac{2}{2^2+1}....\frac{n}{n^2+1}=\lim \frac{1.2.3...n}{(1^2+1)(2^2+1)....+(n^2+1)}\)
\(=\lim \frac{1}{(1+\frac{1}{1})(2+\frac{1}{2})....(n+\frac{1}{n})}\)
Với $n\to +\infty$ thì $(1+\frac{1}{1})(2+\frac{1}{2})....(n+\frac{1}{n})\to +\infty$ nên \(=\lim \frac{1}{(1+\frac{1}{1})(2+\frac{1}{2})....(n+\frac{1}{n})}=0\)
Tính các giới hạn
a) \(lim\dfrac{1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{3}\right)^n}{1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^n}\)
\(lim\left(n^3+n\sqrt{n}-5\right)\)
Giúp mình với ạ
Tìm giới hạn các dãy số sau
a) \(lim\dfrac{2^n+6^n-4^{n-1}}{3^n+6^{n+1}}\)
b) \(lim\dfrac{1+3+5+...+\left(2n+1\right)}{3n^2+4}\)
c) \(lim\dfrac{1+2+3+...+n}{n^2-3}\)
d) \(lim\left[\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\right]\)
e) \(lim\left[\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\right]\)
Tính các giới hạn sau:
\(a.lim\left(\dfrac{\left(n-1\right)!+n!+3}{\left(n+2\right)!-\left(n-2\right)!}\right)\)
b.\(lim\left(\dfrac{2n+1}{n\cdot3^n}\right)\)
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.\(lim\dfrac{3^n-4.2^{n-1}-10}{7.2^n+4^n}\)
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.\(\lim\limits_{n->+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)
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.\(\lim\limits_{n->+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
Tính giới hạn của dãy:
\(lim\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^n}\right)\)
lim\(\dfrac{n^4}{\left(n+1\right)\left(2+n\right)\left(n^3+1\right)}\)
