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

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

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

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

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

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

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 

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

BÀi 1

Để A \(\in\) Z

=>\(\left(n+2\right)⋮\left(n-5\right)\)

=>\([\left(n-5\right)+7]⋮\left(n-5\right)\)

=>\(7⋮\left(n-5\right)\)

=>\(n-5\in\left\{1;7;-1;-7\right\}\)

=>\(n\in\left\{6;13;4;-2\right\}\)

Vậy \(n\in\left\{6;13;4;-2\right\}\)

Giúp mk nha

Arigatou gozaimasu!

a, \(A=\dfrac{5n-4-4n+5}{n-3}=\dfrac{n+1}{n-3}=\dfrac{n-3+4}{n-3}=1+\dfrac{4}{n-3}\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

a.\(A=\dfrac{2n+1}{n-3}+\dfrac{3n-5}{n-3}-\dfrac{4n-5}{n-3}\)

\(A=\dfrac{2n+1+3n-5-4n+5}{n-3}\)

\(A=\dfrac{n+1}{n-3}\)

\(A=\dfrac{n-3}{n-3}+\dfrac{4}{n-3}\)

\(A=1+\dfrac{4}{n-3}\)

Để A nguyên thì \(\dfrac{4}{n-3}\in Z\) hay \(n-3\in U\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

n-3=1 --> n=4

n-3=-1 --> n=2

n-3=2 --> n=5

n-3=-2 --> n=1

n-3=4 --> n=7

n-3=-4 --> n=-1

Vậy \(n=\left\{4;2;5;7;1;-1\right\}\) thì A nhận giá trị nguyên

b.hemm bt lèm:vv

cc

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

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

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

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

\(\lim\limits\left[\sqrt{n}\left(\sqrt{n+1}-n\right)\right]\)

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

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

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

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

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

\(=-\infty\) vì \(\left\{{}\begin{matrix}lim\left(n\sqrt{n}\right)=+\infty\\lim\left(\dfrac{-1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\right)=-\dfrac{1}{1}=-1< 0\end{matrix}\right.\)