Giải bài 10 trang 85 sách bài tập toán 11 - Chân trời sáng tạo tập 1

Đề bài

Tính các giới hạn sau:

a) $\underset{x\to -\mathrm{\infty }}{lim}\left(x^3+2x^2-1\right)$;

b) $\underset{x\to +\mathrm{\infty }}{lim}\frac{x^3+2x^2}{3x^2+1}$;

c) $\underset{x\to -\mathrm{\infty }}{lim}\sqrt{x^2-2x+3}$.

Lời giải chi tiết

a) $\underset{x\to -\mathrm{\infty }}{lim}\left(x^3+2x^2-1\right)$ $=\underset{x\to -\mathrm{\infty }}{lim}\left[x^3\left(1+\frac{2}{x}-\frac{1}{x^3}\right)\right]$

Vì $\underset{x\to -\mathrm{\infty }}{lim}{x}^3$ $=-\mathrm{\infty };\underset{x\to -\mathrm{\infty }}{lim}\left(1+\frac{2}{x}-\frac{1}{x^3}\right)$ $=\underset{x\to -\mathrm{\infty }}{lim}1+\underset{x\to -\mathrm{\infty }}{lim}\frac{2}{x}-\underset{x\to -\mathrm{\infty }}{lim}\frac{1}{x^3}$ $=1>0$

Do đó, $\underset{x\to -\mathrm{\infty }}{lim}\left(x^3+2x^2-1\right)$ $=\underset{x\to -\mathrm{\infty }}{lim}\left[x^3\left(1+\frac{2}{x}-\frac{1}{x^3}\right)\right]$ $=-\mathrm{\infty }$

b) $\underset{x\to +\mathrm{\infty }}{lim}\frac{x^3+2x^2}{3x^2+1}$ $=\underset{x\to +\mathrm{\infty }}{lim}\left[x.\frac{1+\frac{2}{x}}{3+\frac{1}{x^2}}\right]$

Ta có: $\underset{x\to +\mathrm{\infty }}{lim}x$ $=+\mathrm{\infty },\underset{x\to +\mathrm{\infty }}{lim}\frac{1+\frac{2}{x}}{3+\frac{1}{x^2}}$ $=\frac{1+\underset{x\to +\mathrm{\infty }}{lim}\frac{2}{x}}{3+\underset{x\to +\mathrm{\infty }}{lim}\frac{1}{x^2}}$ $=\frac{1}{3}>0$

Do đó, $\underset{x\to +\mathrm{\infty }}{lim}\frac{x^3+2x^2}{3x^2+1}$ $=\underset{x\to +\mathrm{\infty }}{lim}\left[x.\frac{1+\frac{2}{x}}{3+\frac{1}{x^2}}\right]$ $=+\mathrm{\infty }$

c) $\underset{x\to -\mathrm{\infty }}{lim}\sqrt{x^2-2x+3}\underset{x\to -\mathrm{\infty }}{lim}\left[\left|x\right|\sqrt{1-\frac{2}{x}+\frac{3}{x^2}}\right]$ $=\underset{x\to -\mathrm{\infty }}{lim}\left[-x\sqrt{1-\frac{2}{x}+\frac{3}{x^2}}\right]$

Ta có: $\underset{x\to -\mathrm{\infty }}{lim}\left(-x\right)$ $=+\mathrm{\infty };\underset{x\to -\mathrm{\infty }}{lim}\sqrt{1-\frac{2}{x}+\frac{3}{x^2}}$ $=\sqrt{1-\underset{x\to -\mathrm{\infty }}{lim}\frac{2}{x}+\underset{x\to -\mathrm{\infty }}{lim}\frac{3}{x^2}}$ $=1>0$

Do đó, $\underset{x\to -\mathrm{\infty }}{lim}\sqrt{x^2-2x+3}$ $=\underset{x\to -\mathrm{\infty }}{lim}\left[-x\sqrt{1-\frac{2}{x}+\frac{3}{x^2}}\right]$ $=+\mathrm{\infty }$