Giải bài 6 trang 133 – SGK môn Đại số và Giải tích lớp 11

Hướng dẫn:

Áp dụng: $\lim\limits_{x\to \pm\infty }u_nv_n=\lim\limits_{x\to \pm\infty }u_n.\lim\limits_{x\to \pm\infty }v_n$

a)

 $\lim\limits_{x\to +\infty }\,\left( {{x}^{4}}-{{x}^{2}}+x-1 \right)=\lim\limits_{x\to +\infty }\,{{x}^{4}}\left( 1-\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{4}}} \right)\\=\lim\limits_{x\to +\infty }\,{{x}^{4}}.\lim\limits_{x\to +\infty }\,\left( 1-\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{4}}} \right)=+\infty$
Vì $\lim\limits_{x\to +\infty }\,{{x}^{4}}=+\infty$  và $\lim\limits_{x\to +\infty }\,\left( 1-\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{4}}} \right)=1>0$ . 
b) $\lim\limits_{x\to -\infty }\,\left( -2{{x}^{3}}+3{{x}^{2}}-5 \right)=\lim\limits_{x\to -\infty }\,{{x}^{3}}\left( -2+\dfrac{3}{x}-\dfrac{5}{{{x}^{3}}} \right)=\lim\limits_{x\to -\infty }\,{{x}^{3}}.\lim\limits_{x\to -\infty }\,\left( -2+\dfrac{3}{x}-\dfrac{5}{{{x}^{3}}} \right)=+\infty$  
Vì $\lim\limits_{x\to -\infty }\,{{x}^{3}}=-\infty$  và $\lim\limits_{x\to -\infty }\,\left( -2+\dfrac{3}{x}-\dfrac{5}{{{x}^{2}}} \right)=-2<0$
c) $\lim\limits_{x\to -\infty }\,\sqrt{{{x}^{2}}-2x+5}=\lim\limits_{x\to -\infty }\,\sqrt{{{x}^{2}}\left( 1-\dfrac{2}{x}+\dfrac{5}{{{x}^{2}}} \right)}=\lim\limits_{x\to -\infty }\,|x|\sqrt{1-\dfrac{2}{x}+\dfrac{5}{{{x}^{2}}}}=+\infty$
Vì $\lim\limits_{x\to -\infty }\,|x|=+\infty$  và $\lim\limits_{x\to -\infty }\,\sqrt{1-\dfrac{2}{x}+\dfrac{5}{{{x}^{2}}}}=1>0$
d) $\lim\limits_{x\to +\infty }\,\dfrac{\sqrt{{{x}^{2}}+1}+x}{5-2x}=\lim\limits_{x\to +\infty }\,\dfrac{x\left( \sqrt{1+\dfrac{1}{{{x}^{2}}}}+1 \right)}{x\left( \dfrac{5}{x}-2 \right)}=\lim\limits_{x\to +\infty }\,\dfrac{\sqrt{1+\dfrac{1}{{{x}^{2}}}}+1}{\dfrac{5}{x}-2}=-1$ 
Vì $\lim\limits_{x\to \infty }\,\left( \sqrt{1+\dfrac{1}{{{x}^{2}}}}+1 \right)=2$  và $\lim\limits_{x\to +\infty }\,\left( \dfrac{5}{x}-2 \right)=-2$