Giải thích các bước giải:

Câu 1:

$\lim\dfrac{6n^3-2n+1}{n^3-2n}=\lim\dfrac{6-\dfrac{2}{n^2}+\dfrac{1}{n^3}}{1-\dfrac{2}{n^2}}=\dfrac{6}{1}=6$

Câu 2:

$\lim\dfrac{1-n+2n^2}{5n^2+n}=\lim\dfrac{\dfrac{1}{n^2}-\dfrac{1}{n}+2}{\dfrac{5}{n^2}+\dfrac{1}{n}}=\dfrac{2}{0}=+\infty$

Câu 3:

$\lim\dfrac{2n^3-4n^2+3n+3}{n^3-5n+7}=\lim\dfrac{2-\dfrac{4}{n}+\dfrac{3}{n^2}+\dfrac{3}{n^3}}{1-\dfrac{5}{n^2}+\dfrac{7}{n^3}}=\dfrac{2}{1}=2$

Câu 4:

$\lim\dfrac{-2n^2+n+2}{3n^4+5}=\lim\dfrac{-\dfrac{2}{n^2}+\dfrac{1}{n^3}+\dfrac{2}{n^4}}{3+\dfrac{5}{n^4}}=\dfrac{0}{3}=0$

Câu 5:

$\lim\dfrac{n^2+4n-5}{3n^3+n^2+7}=\lim\dfrac{\dfrac{1}{n}+\dfrac{4}{n^2}-\dfrac{5}{n^3}}{3+\dfrac{1}{n}+\dfrac{7}{n^3}}=\dfrac{0}{3}=0$

Câu 6:

$\lim \dfrac{n^5+n^4-n-2}{4n^3+6n^2+9}=\lim \dfrac{1+\dfrac{1}{n}-\dfrac{1}{n^4}-\dfrac{2}{n^5}}{\dfrac{4}{n^2}+\dfrac{6}{n^3}+\dfrac{9}{n^5}}=\dfrac{1}{0}=+\infty$

Câu 7:

$\lim \dfrac{7n^2-3n+2}{n^2+5}=\lim \dfrac{7-\dfrac{3}{n}+\dfrac{2}{n^2}}{1+\dfrac{5}{n^2}}=\dfrac{7}{1}=7$

Câu 8:

$\lim \dfrac{8n^3+2n-1}{2n^2-n}=\lim \dfrac{8+\dfrac{2}{n^2}-\dfrac{1}{n^3}}{\dfrac{2}{n}-\dfrac{1}{n^2}}=\dfrac{3}{0}=+\infty$

Câu 9:

$\lim (\dfrac{2n^3}{2n^2+3}+\dfrac{1-5n^2}{5n+1})$

$=\lim\dfrac{10n^4+2n^3+2n^2+3-10n^4-15n^2}{10n^3+2n^2+15n+3}$

$=\lim\dfrac{2n^3-13n^2+3}{10n^3+2n^2+15n+3}$

$=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{10+\dfrac{2}{n}+\dfrac{15}{n^2}+\dfrac{3}{n^3}}$

$=\dfrac{2}{10}=\dfrac{1}{5}$

Câu 10:

$\lim \dfrac{-3n^5+7n^3-11}{n^5+n^4-3n}=\lim \dfrac{-3+\dfrac{7}{n^2}-\dfrac{11}{n^5}}{1+\dfrac{1}{n}-\dfrac{3}{n^4}}=\dfrac{-5}{1}=-5$

Câu 11:

$\lim \dfrac{2n^2-3}{n^6+5n^5}=\lim \dfrac{\dfrac{2}{n^4}-\dfrac{3}{n^6}}{1+\dfrac{5}{n}}=\dfrac{0}{1}=0$

Câu 12:

$\lim\dfrac{\sqrt{2n^2-n}}{1-3n^2}$ 

$=\lim\dfrac{\sqrt{2n^2-n}:n^2}{\dfrac{1}{n^2}-3}$ 

$=\lim\dfrac{\sqrt{\dfrac{2}{n^2}-\dfrac{1}{n^3}}}{\dfrac{1}{n^2}-3}$ 

$=\dfrac{\sqrt{0-0}}{0-3}$ 

$=0$ 

Câu 13:

$\lim \dfrac{\sqrt[3]{n^3+n}}{n+2}=\lim \dfrac{\sqrt[3]{1+\dfrac{1}{n^2}}}{1+\dfrac{2}n}=\dfrac{1}{1}=1$

Câu 14:

$\lim\dfrac{\sqrt{2n^4+3n-2}}{2n^2-n+3}=\lim\dfrac{\sqrt{2+\dfrac{3}{n^3}-\dfrac 2{n^4}}}{2-\dfrac 1n+\dfrac 3{n^2}}=\dfrac{\sqrt2}{2}$

Câu 15:

$\lim\dfrac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\dfrac{\sqrt[3]{1-\dfrac{7}{n^3}-\dfrac{5}{n^5}+\dfrac{8}{n^6}}}{\dfrac{1}{n}+\dfrac{12}{n^2}}=\dfrac{1}{0}=+\infty$

Câu 16:

$\lim \dfrac{\sqrt{n^2+1}-\sqrt{n+1}}{3n+2}=\lim \dfrac{\sqrt{1+\dfrac{1}{n^2}}-\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}}{3+\dfrac{2}{n}}=\dfrac{1-0}{3}=\dfrac 13$

Câu 17:

$\lim (3n^3-7n+11)=\lim n^3(3-\dfrac{7}{n^2}+\dfrac{11}{n^3})=+\infty.3=+\infty$

Câu 18:
$\lim \sqrt{2n^4-n^2+n+2}=\lim n^2.\sqrt{2-\dfrac{1}{n^2}+\dfrac{1}{n^3}+\dfrac{2}{n^4}}=+\infty.\sqrt{2}=+\infty$

Câu 19:

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

Câu 20:

$\lim \dfrac{1+2+..+n}{n^2}=\lim\dfrac{\dfrac{n(n+1)}{2}}{n^2}=\lim \dfrac{n^2+n}{2n^2}=\lim\dfrac{1}{2}+\dfrac{1}{2n}=\dfrac 12$

Câu 21:

$\lim\dfrac{n\sqrt{2+4+..+2n}}{3n^2+n-2}$

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

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

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

$=\lim\dfrac{n\sqrt{n^2+n}}{3n^2+n-2}$

$=\lim\dfrac{\sqrt{1+\dfrac 1n}}{3+\dfrac 1n-\dfrac{2}{n^2}}$

$=\dfrac{1}{3}$

Câu 22:

$\lim\dfrac{1^3+2^3+..+n^3}{n^4+n^3+3n+2}$

$=\lim\dfrac{(1+2+..+n)^2}{n^4+n^3+3n+2}$

$=\lim\dfrac{(\dfrac{n(n+1)}{2})^2}{n^4+n^3+3n+2}$

$=\lim\dfrac 14\dfrac{n^2(n+1)^2}{n^4+n^3+3n+2}$

$=\lim\dfrac 14\dfrac{(1+\dfrac 1n)^2}{1+\dfrac 1n+\dfrac{3}{n^3}+\dfrac{2}{n^4}}$
$=\dfrac 14.\dfrac{1}{1}=\dfrac 14$

Câu 23:

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

$=\lim \dfrac{n\sqrt{\dfrac{(2n-1+1)n}{2}}}{2n^2+n+1}$

$=\lim \dfrac{n\sqrt{n^2}}{2n^2+n+1}$

$=\lim \dfrac{n^2}{2n^2+n+1}$

$=\lim\dfrac{1}{2+\dfrac 1n+\dfrac 1{n^2}}$

$=\dfrac 12$