1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-6n^5+3n^3-1}{n^4-8n}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-5n^7+8n^5-n}{5n^6-2n}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-6n^5+3n^3-1}{n^4-8n}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-5n^7+8n^5-n}{5n^6-2n}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{6-\dfrac{8}{n}}{1-\dfrac{1}{n}}=\dfrac{6-0}{1-0}\)
\(=\dfrac{6}{1}=6\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\left(\dfrac{1}{n}\cdot\dfrac{1+\dfrac{5}{n}-\dfrac{3}{n^2}}{\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\right)\)
=0
Tìm giới hạn dãy số :
\(a,lim\dfrac{5n+1}{2n}\\ b,lim\dfrac{6n^2+8n+1}{5n^2+3}\\ c,lim\dfrac{3^n+2^n}{4.3^n}\\ d,lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}\)
a: \(\lim\limits\dfrac{5n+1}{2n}=\lim\limits\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=\lim\limits\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5+0}{2}=\dfrac{5}{2}\)
b: \(\lim\limits\dfrac{6n^2+8n+1}{5n^2+3}\)
\(=\lim\limits\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}\)
\(=\lim\limits\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}\)
\(=\dfrac{6+0+0}{5+0}=\dfrac{6}{5}\)
c: \(\lim\limits\dfrac{3^n+2^n}{4\cdot3^n}\)
\(=\lim\limits\dfrac{\dfrac{3^n}{3^n}+\left(\dfrac{2}{3}\right)^n}{4\cdot\left(\dfrac{3^n}{3^n}\right)}\)
\(=\lim\limits\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1+0}{4}=\dfrac{1}{4}\)
d: \(\lim\limits\dfrac{\sqrt{n^2+5n+3}}{6n+2}\)
\(=\lim\limits\dfrac{\sqrt{\dfrac{n^2}{n^2}+\dfrac{5n}{n^2}+\dfrac{3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}\)
\(=\lim\limits\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}\)
\(=\dfrac{\sqrt{1+0+0}}{6}=\dfrac{1}{6}\)
\(a,lim\dfrac{5n+1}{2n}=lim\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=lim\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5}{2}\\ b,lim\dfrac{6n^2+8n+1}{5n^2+3}=lim\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}=lim\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}=\dfrac{6}{5}\)
\(c,lim\dfrac{3^n+2^n}{4.3^n}=\dfrac{\dfrac{3^n}{3^n}+\dfrac{2^n}{3^n}}{\dfrac{4.3^n}{3^n}}=\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1}{4}\)
\(d,lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}=lim\dfrac{\sqrt{\dfrac{n^2+5n+3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}=lim\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}=\dfrac{1}{6}\)
\(a\text{)}lim\dfrac{5n+1}{2n}=lim\dfrac{5}{2}+lim\dfrac{1}{2n}=\dfrac{5}{2}\)
\(b\text{)}lim\dfrac{6n^2+8n+1}{5n^2+3}=lim\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}=\dfrac{6}{5}\)
\(c\text{)}lim\dfrac{3^n+2^n}{4.3^n}=lim\dfrac{\left(\dfrac{3}{3}\right)^n+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1}{4}\)
\(d\text{)}lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}=lim\dfrac{n\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{n\left(6+\dfrac{2}{n}\right)}=lim\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}=\dfrac{1}{6}\)
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) \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}=\lim\limits_{n\rightarrow\infty}\dfrac{2n\left(1-\dfrac{4}{n}\right)}{n\left(1-\dfrac{1}{n}\right)}=2\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}=\dfrac{1}{4n}=\infty\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}\)
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n\left(-1+\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\left[n\left(\dfrac{3+\dfrac{5}{n}-\dfrac{3}{n^2}}{-1+\dfrac{5}{n}}\right)\right]\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{5}{n}-\dfrac{3}{n^2}}{-1+\dfrac{5}{n}}=\dfrac{3+0-0}{-1+0}=\dfrac{3}{-1}=-3< 0\end{matrix}\right.\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{7n^2-4}{n-5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(7-\dfrac{4}{n^2}\right)}{n\left(1-\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\left[n\cdot\dfrac{\left(7-\dfrac{4}{n^2}\right)}{1-\dfrac{5}{n}}\right]\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{7-\dfrac{4}{n^2}}{1-\dfrac{5}{n}}=\dfrac{7-0}{1-0}=7>0\end{matrix}\right.\)
Tính :6/ lim\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
7/ lim \(\dfrac{\sqrt{n^3-2n+5}}{3+5n}\)
10/ lim\(\dfrac{1+3+5+...+\left(2n+1\right)}{3n^3+4}\)
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\(\dfrac{\sqrt{n^2-4n}-\sqrt{4n^2+1}}{\sqrt{3n^2+1}-n}\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
tính
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}\)
3) \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(3n+5-\dfrac{3}{n}\right)}{-n\left(1-\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n+5-\dfrac{3}{n}}{-\left(1-\dfrac{5}{n}\right)}\)
\(=\left[{}\begin{matrix}-\infty\left(n\rightarrow+\infty\right)\\+\infty\left(n\rightarrow-\infty\right)\end{matrix}\right.\)
Bài 2,3 tương tự, bạn tự làm nhé!
Tính các giới hạn sau:
a) \(\lim\limits\dfrac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}\)
b) \(\lim\limits\dfrac{1}{\sqrt{3n+2}-\sqrt{2n+1}}\)
c) \(\lim\limits\dfrac{4.3^n+7^{n+1}}{2.5^n+7^n}\)
a.
\(A=\lim\frac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\lim \frac{\sqrt[3]{\frac{n^6-7n^3-5n+8}{n^3}}}{\frac{n+12}{n}}=\lim \frac{\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}}{1+\frac{12}{n}}\)
Ta thấy:
\(\lim\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}=\infty \)
\(\lim (1+\frac{12}{n})=1\)
Suy ra $A=\infty$
b.
\(B=\lim\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=\lim \frac{1}{\frac{3n+2-(2n+1)}{\sqrt{3n+2}+\sqrt{2n+1}}}=\lim \frac{\sqrt{3n+2}+\sqrt{2n+1}}{n+1}\)
\(=\lim \frac{\sqrt{\frac{3n+2}{n}}+\sqrt{\frac{2n+1}{n}}}{\frac{n+1}{\sqrt{n}}}=\lim \frac{\sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}}}{\sqrt{n}+\frac{1}{\sqrt{n}}}\)
Ta thấy:
\(\lim( \sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}})=\sqrt{3}+\sqrt{2}>0\)
\(\lim (\sqrt{n}+\frac{1}{\sqrt{n}})=\infty\)
$\Rightarrow B=\infty$
c.
\(C=\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{4(\frac{3}{7})^n+7}{2(\frac{5}{7})^n+1}\)
Ta thấy:
\(\lim [4(\frac{3}{7})^n+7]=4.0+7=7\) với $|\frac{3}{7}|<1$
\(\lim [2(\frac{5}{7})^n+1]=2.0+1=1\) với $|\frac{5}{7}|<1$
$\Rightarrow C=\frac{7}{1}=7$
Tìm các giới hạn sau:
\(a,lim\dfrac{2n+1}{-3n+2}\)
\(b,lim\dfrac{5n^3-2n+1}{n-2n^3}\)
\(a,lim\dfrac{2n+1}{-3n+2}\)
\(=lim\dfrac{2+\dfrac{1}{n}}{-3+\dfrac{2}{n}}=-\dfrac{2}{3}\)
\(b,lim\dfrac{5n^3-2n+1}{n-2n^3}\)
\(=lim\dfrac{5-\dfrac{2}{n^2}+\dfrac{1}{n^3}}{\dfrac{1}{n^2}-2}=\dfrac{5}{-2}\)