Tính các giới hạn sau
1. $\underset{n\to +\infty }{\lim }\frac{2n+1}{5n+9}$.
2. $\underset{n\to +\infty }{\lim }\frac{n-7}{n^2+9n+2}$.
3. $\underset{n\to +\infty }{\lim }\frac{n^3-3n+1}{(2n+3)\left(n^2+7\right)}$.
4. $\underset{n\to +\infty }{\lim }\frac{n^2-9n+2}{5n+9}$.
5. $\underset{n\to +\infty }{\lim }\frac{3^n-2^n}{3^{n+2}+1}$.
6. $\underset{n\to +\infty }{\lim }\frac{2^{2n+1}-7^n}{2^{3n-1}+7^{n+1}}$.
7. $\underset{n\to +\infty }{\lim }\left(-2n^3+9n^2+1\right)$.
8. $\underset{n\to +\infty }{\lim }\left(\sqrt{4n^2+2}-5n\right)$.
9. $\underset{n\to +\infty }{\lim }\left(\sqrt{n^2+1}-\sqrt{n^2+8n+2}\right)$.

(AI giải đề cho tôi)
Bài 15 – Tính các giới hạn dãy số:

1) $\underset{n\to +\infty }{\lim }\frac{2n+1}{5n+9}=\lim \frac{2+\frac{1}{n}}{5+\frac{9}{n}}=\frac{2}{5}$.

2) $\underset{n\to +\infty }{\lim }\frac{n-7}{n^2+9n+2}=\lim \frac{\frac{1}{n}-\frac{7}{n^2}}{1+\frac{9}{n}+\frac{2}{n^2}}=0$.

3) Chia tử và mẫu cho $n^3$: $\lim \frac{1-\frac{3}{n^2}+\frac{1}{n^3}}{\left(2+\frac{3}{n}\right)\left(1+\frac{7}{n^2}\right)}=\frac{1}{2\cdot 1}=\frac{1}{2}$.

4) $\underset{n\to +\infty }{\lim }\frac{n^2-9n+2}{5n+9}=\lim \frac{n-9+\frac{2}{n}}{5+\frac{9}{n}}=+\infty$.

5) Chia tử và mẫu cho $3^n$: $\lim \frac{1-{\left(\frac{2}{3}\right)}^n}{9+\frac{1}{3^n}}=\frac{1-0}{9+0}=\frac{1}{9}$.

6) $\frac{2^{2n+1}-7^n}{2^{3n-1}+7^{n+1}}=\frac{2\cdot 4^n-7^n}{\frac{1}{2}\cdot 8^n+7\cdot 7^n}$. Chia tử và mẫu cho $8^n$ (số mũ lớn nhất): $\lim \frac{2{\left(\frac{1}{2}\right)}^n-{\left(\frac{7}{8}\right)}^n}{\frac{1}{2}+7{\left(\frac{7}{8}\right)}^n}=\frac{0-0}{\frac{1}{2}+0}=0$.

7) Vì $-2n^3+9n^2+1=n^3\left(-2+\frac{9}{n}+\frac{1}{n^3}\right)$ và $-2+\frac{9}{n}+\frac{1}{n^3}\to -2<0$, nên $\underset{n\to +\infty }{\lim }\left(-2n^3+9n^2+1\right)=-\infty$.

8) Nhân liên hợp: $\sqrt{4n^2+2}-5n=\frac{4n^2+2-25n^2}{\sqrt{4n^2+2}+5n}=\frac{-21n^2+2}{\sqrt{4n^2+2}+5n}$. Chia tử và mẫu cho $n$: $\lim \frac{-21n+\frac{2}{n}}{\sqrt{4+\frac{2}{n^2}}+5}=-\infty$.

9) Nhân liên hợp: $\sqrt{n^2+1}-\sqrt{n^2+8n+2}=\frac{(n^2+1)-(n^2+8n+2)}{\sqrt{n^2+1}+\sqrt{n^2+8n+2}}=\frac{-8n-1}{\sqrt{n^2+1}+\sqrt{n^2+8n+2}}$. Chia tử và mẫu cho $n$: $\lim \frac{-8-\frac{1}{n}}{\sqrt{1+\frac{1}{n^2}}+\sqrt{1+\frac{8}{n}+\frac{2}{n^2}}}=\frac{-8}{1+1}=-4$.