Tính: a) lim (2n - 3n^3 + 1)/(n^3 + n^2) b) lim (2n căn n)/(n^2 + 2n - 1) c) lim (3n^3 - 5n +1)/(n^2 + 4) d) lim ((2 + 3n) ^3 (n + 1)^2)/(1 - 4n^5) e) lim (căn (n^2 + n - 1) - căn (4n^2 - 2))/(n + 3) f) lim (n^2 + 2n - 5) h) lim (-n^2 + n căn n + 1) k) lim (căn (n^2 + n) - căn (n^2 - 1)) Các bạn giúp mình lm bài tập này với ạ!

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

b) $\lim\dfrac{2n\sqrt n}{n^2+2n-1}=\lim\dfrac{\dfrac{2}{\sqrt n}}{1+\dfrac{2}{n}-\dfrac{1}{n^2}}=0$

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

d) $\lim\dfrac{(2+3n)^3(n+1)^2}{1-4n^5}=\lim\dfrac{(\dfrac{2}{n}+3)^3(1+\dfrac{1}{n})^2}{\dfrac{1}{n^5}-4}=\dfrac{3^3.1^2}{-4}=\dfrac{-27}{4}$

e) $\lim\dfrac{\sqrt{n^2+n-1}-\sqrt{4n^2-2}}{n+3}=\lim\dfrac{\sqrt{1+\dfrac{1}{n}-\dfrac{1}{n^2}}-\sqrt{4-\dfrac{2}{n^2}}}{1+\dfrac{3}{n}}=\dfrac{1-\sqrt4}{1}=-1$

f) $\lim(n^2+2n-5)=\lim[n^2(1+\dfrac{2}{n}-\dfrac{5}{n^2})]=+\infty$

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

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

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

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