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

3) $\underset{n\to +\infty }{\lim }\frac{n^3-3n+1}{(2n+3)(n^2+7)}=\lim \frac{1-\frac{3}{n^2}+\frac{1}{n^3}}{\left(2+\frac{3}{n}\right)\left(1+\frac{7}{n^2}\right)}=\frac{1}{2\cdot 1}=\frac{1}{2}$.

4) $\underset{n\to +\infty }{\lim }\frac{n^2-9n+2}{5n+9}=\lim \frac{n-9+\frac{2}{n}}{5+\frac{9}{n}}=+\infty$ (tử cấp $n^2$ trội hơn mẫu cấp $n$).

5) $\underset{n\to +\infty }{\lim }\frac{3^n-2^n}{3^{n+2}+1}=\lim \frac{1-{\left(\frac{2}{3}\right)}^n}{9+\frac{1}{3^n}}=\frac{1}{9}$ (do ${\left(\frac{2}{3}\right)}^n\to 0$, $\frac{1}{3^n}\to 0$).

6) $\underset{n\to +\infty }{\lim }\frac{2^{2n+1}-7^n}{2^{3n-1}+7^{n+1}}$. Chia tử và mẫu cho $8^n=2^{3n}$ (cấp cao nhất ở mẫu): $\frac{2^{2n+1}}{8^n}=2{\left(\frac{1}{2}\right)}^n\to 0$; $\frac{7^n}{8^n}={\left(\frac{7}{8}\right)}^n\to 0$; $\frac{2^{3n-1}}{8^n}=\frac{1}{2}$; $\frac{7^{n+1}}{8^n}=7{\left(\frac{7}{8}\right)}^n\to 0$. Do đó giới hạn bằng $\frac{0-0}{\frac{1}{2}+0}=0$.

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

8) $\underset{n\to +\infty }{\lim }\left(\sqrt{4n^2+2}-5n\right)=\lim \frac{4n^2+2-25n^2}{\sqrt{4n^2+2}+5n}=\lim \frac{-21n^2+2}{\sqrt{4n^2+2}+5n}=-\infty$ (tử cấp $n^2$ hệ số âm, mẫu cấp $n$).

9) Nhân liên hợp: $\underset{n\to +\infty }{\lim }\left(\sqrt{n^2+1}-\sqrt{n^2+8n+2}\right)=\lim \frac{-8n-1}{\sqrt{n^2+1}+\sqrt{n^2+8n+2}}=\lim \frac{-8-\frac{1}{n}}{\sqrt{1+\frac{1}{n^2}}+\sqrt{1+\frac{8}{n}+\frac{2}{n^2}}}=\frac{-8}{2}=-4$.