Giải Thực hành 4 trang 76 Toán 11 tập 1 Chân trời

a) $\lim_{x\rightarrow +\infty }\frac{1-3x^{2}}{x^{2}+2x}$

= $\lim_{x\rightarrow +\infty }\frac{1-3x^{2}}{x^{2}+2x} = \lim_{x\rightarrow +\infty }\frac{\frac{1}{x^{2}}-3}{1+\frac{2}{x}}$

= $\frac{\lim_{x\rightarrow +\infty }\left (\frac{1}{x^{2}}-3 \right )}{\lim_{x\rightarrow +\infty }\left (1+\frac{2}{x}  \right )}$

= $\frac{\lim_{x\rightarrow +\infty }\frac{1}{x^{2}}-3}{1+\lim_{x\rightarrow +\infty }\frac{2}{x}}$

= $\frac{0-3}{1+0} = -3$

b) $\lim_{x\rightarrow -\infty }\frac{2}{x+1}$

= $\lim_{x\rightarrow -\infty}\frac{\frac{2}{x}}{1+\frac{1}{x}}$

= $\frac{lim_{x\rightarrow -\infty}\frac{2}{x}}{\lim_{x\rightarrow -\infty}\left ( 1+\frac{1}{x} \right )} $

= $\frac{lim_{x\rightarrow -\infty}\frac{2}{x}}{1+lim_{x\rightarrow -\infty}\frac{1}{x}}$ 

= $\frac{0}{1+0} = 0$