\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)

\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)

\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)

\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)

\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)

\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)

\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)

\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)

Lim 3.4n-2.13n/5n+6.13n

a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả

b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)

c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)

d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)

e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)

f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)

g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)

\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)

\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)

a) lim \(\left(-3n^3+n^2-1\right)\)

minh le oi ban dao mau so cua ban len cho tu uong roi thay vi tri cua mau thanh n³ +2n

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 $|\frac{5}{7}|<1$

$\Rightarrow C=\frac{7}{1}=7$

\(a=\lim\dfrac{-2n^2}{\sqrt{n^2+2}+\sqrt{n^2+4}}=\lim\dfrac{-2n}{\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1+\dfrac{4}{n^2}}}=\dfrac{-\infty}{2}=-\infty\)

\(b=\lim\dfrac{3-5n^2+10n}{n-2}=\lim\dfrac{-5n+10+\dfrac{3}{n}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)

\(c=\lim\left(\dfrac{1-\dfrac{1}{n}}{\dfrac{\sqrt{3}}{n}-1}-4.2^n\right)=-1-\infty=-\infty\)

\(d=\lim\dfrac{n^3-4n-\left(3n^2+4\right)\left(n-2\right)}{n^2-2n}=\lim\dfrac{-2n^3+6n^2-8n+8}{n^2-2n}\)

\(\lim\dfrac{-2n+6-\dfrac{8}{n}+\dfrac{8}{n^2}}{1-\dfrac{2}{n}}=\dfrac{-\infty}{1}=-\infty\)

\(e=\lim\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt{5}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{5}}=\dfrac{1-\sqrt{5}}{1+\sqrt{5}}\)

a. ĐKXĐ: \(n\ge0\)

\(lim_{n\rightarrow0}\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=\dfrac{\sqrt{2.0+1}}{\sqrt{8.0}+1}=1\)

\(lim_{n\rightarrow+\infty}\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=lim_{n\rightarrow+\infty}\dfrac{\sqrt{2+\dfrac{1}{n}}}{\sqrt{8}+\dfrac{1}{\sqrt{n}}}=\dfrac{1}{2}\)

b. ĐKXĐ: \(\left\{{}\begin{matrix}n\ne0\\n\le\dfrac{-1-\sqrt{21}}{2}\\n\ge\dfrac{-1+\sqrt{21}}{2}\end{matrix}\right.\)

\(lim_{n\rightarrow+\infty}\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\)\(lim_{n\rightarrow+\infty}\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=-2\)

\(lim_{n\rightarrow-\infty}\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\)\(lim_{n\rightarrow-\infty}\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=-1\)

a, \(lim\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=lim\dfrac{\sqrt{n}.\sqrt{2+\dfrac{1}{n}}}{\sqrt{n}\left(\sqrt{8}+\dfrac{1}{n}\right)}=\dfrac{\sqrt{2}}{\sqrt{8}}=\dfrac{1}{2}\)

b, \(lim\dfrac{3n+\sqrt{n^2+n-5}}{-2n}\)

\(=lim\left(\dfrac{3}{2}-\dfrac{\sqrt{n^2+n-5}}{2n}\right)\)

\(=lim\left(\dfrac{3}{2}-\dfrac{n\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{2n}\right)=\dfrac{3}{2}-\dfrac{1}{2}=1\)

\(\lim\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=\lim\dfrac{\sqrt{n}.\sqrt{2+\dfrac{1}{n}}}{\sqrt{n}\left(\sqrt{8}+\dfrac{1}{\sqrt{n}}\right)}=\lim\dfrac{\sqrt{2+\dfrac{1}{n}}}{\sqrt{8}+\dfrac{1}{\sqrt{n}}}=\dfrac{\sqrt{2}}{\sqrt{8}}=\dfrac{1}{2}\)

\(\lim\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\lim\dfrac{n\left(3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}\right)}{-2n}=\lim\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=\dfrac{3+1}{-2}=-2\)

Câu a:

\(\lim\dfrac{n\sqrt{1+2+...+2n}}{3n^2+n-2}=\lim\dfrac{n\sqrt{\dfrac{2n\left(2n+1\right)}{2}}}{3n^2+n-2}=\lim\dfrac{\sqrt{2+\dfrac{1}{n}}}{3+\dfrac{1}{n}-\dfrac{2}{n^2}}=\dfrac{\sqrt{2}}{3}\)