Cau 1: B= lim 4n^2+3n+1/ (3n-1)^2 = 4/9
Cau 2: lim( can n^2+2n - can n^2 -2n =0
Cau 1: B= lim 4n^2+3n+1/ (3n-1)^2 = 4/9
Cau 2: lim( can n^2+2n - can n^2 -2n =0
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 \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^3+3n^2-1}{n^2-2n}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2-1}{-2n+3}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^5+3n^3-1}{n^3-2n}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^7+3n^5-n}{3n^2-2n}\)
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 \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\)
tinh gioi han \(\lim\dfrac{1+cos\left(n^2\right)}{1+2n}\)
Giúp mình giải với
a) lim \(\frac{\sqrt{n^2-4n}-\sqrt{4n+1}}{\sqrt{3n^2+1}+n}\)
b)lim \(\frac{\sqrt[3]{8n^3+n^2}-n}{2n-3}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4x^4-3n^2+4\right)\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2-n+2}{n^3+2n^2-3}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n+2}{3n^3-2n+n^2}\)
Tính giới hạn: \(lim\left(\dfrac{2n^2+3n}{n+1}-\dfrac{2n^3-3}{n^2-1}\right)\)