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

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

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

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

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

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

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

Dang này thì cứ chọn số hạng có mũ cao nhất trên tử và mẫu là được. Nó là ngắt vô cùng lớn hay bé gì đấy

\(=lim\dfrac{8n^6}{3n^6}=\dfrac{8}{3}\)

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

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

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

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

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

1: \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)\)

\(=\lim\limits_{n\rightarrow\infty}\left[n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^3}+\dfrac{4}{n^5}\right)\right]\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n^5=+\infty\\\lim\limits_{n\rightarrow\infty}\left(-2+\dfrac{4}{n}-\dfrac{3}{n^3}+\dfrac{4}{n^5}\right)=-2< 0\end{matrix}\right.\)

2: \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-3+\dfrac{2}{n^2}\right)}{n\left(1-\dfrac{2}{n}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(-3+\dfrac{2}{n^2}\right)}{1-\dfrac{2}{n}}\)

\(=-\infty\) vì \(\lim\limits_{n\rightarrow\infty}n=+\infty;\lim\limits_{n\rightarrow\infty}\dfrac{-3+\dfrac{2}{n^2}}{1-\dfrac{2}{n}}=-\dfrac{3}{1}=-3< 0\)

1.

\(\lim (n^3+4n^2-1)=\infty\) khi $n\to \infty$

2. 

\(\lim\limits_{n\to -\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to -\infty}\frac{-\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to -\infty}\frac{-(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{-1}{3}\)

\(\lim\limits_{n\to +\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to +\infty}\frac{\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to +\infty}\frac{(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{1}{3}\)

3.

\(\lim \frac{1+2+...+n}{2n^2}=\lim \frac{n(n+1)}{4n^2}=\lim \frac{n^2+n}{4n^2}\\ =\lim (\frac{1}{4}+\frac{1}{4n})=\frac{1}{4}\)

4.

\(\lim \frac{3^n-4.2^{n-1}-10}{7.2^n+4^n}=\lim \frac{(\frac{3}{4})^n-(\frac{2}{4})^{n-1}-\frac{10}{4^n}}{7(\frac{2}{4})^n+1}\\ =\lim \frac{(\frac{3}{4})^n-(\frac{1}{2})^{n-1}-\frac{10}{4^n}}{7(\frac{1}{2})^n+1}\\ =\frac{0-0-0}{7.0+1}=0\)

Lời giải:

\(\lim\limits \frac{4n^2+3n+1}{(3n-1)^2}=\lim\limits\frac{4n^2+3n+1}{9n^2-6n+1}=\lim\limits\frac{4+\frac{3}{n}+\frac{1}{n^2}}{9-\frac{6}{n}+\frac{1}{n^2}}=\frac{4}{9}\)

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

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

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

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

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

\(=\lim\limits\dfrac{1-\dfrac{3}{n}+\dfrac{3}{n^2}}{1-\dfrac{1}{n^2}}=\dfrac{1-0+0}{1-0}=1\)