\(0\le1+cosn^2\le2\Rightarrow0\le\dfrac{1+cosn^2}{1+2n}\le\dfrac{2}{1+2n}\)
Mà \(\lim\left(0\right)=\lim\left(\dfrac{2}{1+2n}\right)=0\)
\(\Rightarrow\lim\dfrac{1+cosn^2}{1+2n}=0\)
bai nay la gioi han cua day so nhe
Cau 1: B= lim 4n^2+3n+1/ (3n-1)^2 = ?
Cau 2: lim( can n^2+2n - can n^2 -2n =?
Tính giới hạn: \(lim\left(\dfrac{2n^2+3n}{n+1}-\dfrac{2n^3-3}{n^2-1}\right)\)
Tính giới hạn: \(lim\left(\dfrac{n+1}{n^2+2n}-\dfrac{1}{n-1}\right)\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4x^4-3n^2+4\right)\)
1) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\)
3) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\)
1) tính \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4x^4-3n^2+4\right)\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)
\(lim_{x->-\infty}\left(\sqrt{x^2+1}+x-1\right)\\ lim\dfrac{\sqrt{4n^2+n-1}+n}{\sqrt{n^4+2n^3-1}-n}\)
tính lim của lim\(\frac{4n^5-n+1}{\left(2n+1\right)\left(-n+1\right)\left(n^2+2\right)}\)
\(\lim\limits_{\rightarrow}\left(\dfrac{1}{1.3}+\dfrac{1}{2.4}+...+\dfrac{1}{n\left(n+2\right)}\right)\)
