Giải bài 4.24 trang 165 - SBT Đại số và Giải tích lớp 11

a)

$\begin{align} & \lim_{x\to -\infty }\limits\,f\left( x \right)=\lim_{x\to -\infty }\limits\,\dfrac{\sqrt{{{x}^{2}}-3x}}{x+2} \\ & =\lim_{x\to -\infty }\limits\,\dfrac{-x\sqrt{1-\dfrac{3}{x}}}{x\left( 1+\dfrac{2}{x} \right)}=\lim_{x\to -\infty }\limits\,\dfrac{-\sqrt{1-\dfrac{3}{x}}}{1+\dfrac{2}{x}}=-1 \\ \end{align}$

$\begin{align} & \lim_{x\to +\infty }\limits\,f\left( x \right)=\lim_{x\to +\infty }\limits\,\dfrac{\sqrt{{{x}^{2}}-3x}}{x+2} \\ & =\lim_{x\to +\infty }\limits\,\dfrac{x\sqrt{1-\dfrac{3}{x}}}{x\left( 1+\dfrac{2}{x} \right)}=\lim_{x\to +\infty }\limits\,\dfrac{\sqrt{1-\dfrac{3}{x}}}{1+\dfrac{2}{x}}=1 \\ \end{align}$

b)

$\begin{align} & \lim_{x\to -\infty }\limits\,\left( x+\sqrt{{{x}^{2}}-x+1} \right)=\lim_{x\to -\infty }\limits\,\dfrac{x-1}{x-\sqrt{{{x}^{2}}-x+1}} \\ & =\lim_{x\to -\infty }\limits\,\dfrac{x\left( 1-\dfrac{1}{x} \right)}{x\left( 1+\sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}} \right)} \\ & =\lim_{x\to -\infty }\limits\,\dfrac{1-\dfrac{1}{x}}{1+\sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}}}=\dfrac{1}{2} \\ \end{align}$

$\lim_{x\to +\infty }\limits\,\left( x+\sqrt{{{x}^{2}}-x+1} \right)=\lim_{x\to +\infty }\limits\,x\left( 1+\sqrt{1-\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}} \right)=+\infty$  

c)

$\begin{align} & \lim_{x\to -\infty }\limits\,\left( \sqrt{{{x}^{2}}-x}-\sqrt{{{x}^{2}}+1} \right)=\lim_{x\to -\infty }\limits\,\dfrac{-x-1}{\sqrt{{{x}^{2}}-x}+\sqrt{{{x}^{2}}+1}} \\ & =\lim_{x\to -\infty }\limits\,\dfrac{x\left( -1-\dfrac{1}{x} \right)}{-x\left( \sqrt{1-\dfrac{1}{x}}+\sqrt{1+\dfrac{1}{x}} \right)} \\ & =\lim_{x\to -\infty }\limits\,\dfrac{-1-\dfrac{1}{x}}{-\left( \sqrt{1-\dfrac{1}{x}}+\sqrt{1+\dfrac{1}{x}} \right)}=\dfrac{1} 2 \\ \end{align}$

$\begin{align} & \lim_{x\to +\infty }\limits\,\left( \sqrt{{{x}^{2}}-x}-\sqrt{{{x}^{2}}+1} \right)=\lim_{x\to +\infty }\limits\,\dfrac{-x-1}{\sqrt{{{x}^{2}}-x}+\sqrt{{{x}^{2}}+1}} \\ & =\lim_{x\to +\infty }\limits\,\dfrac{x\left( -1-\dfrac{1}{x} \right)}{x\left( \sqrt{1-\dfrac{1}{x}}+\sqrt{1+\dfrac{1}{x}} \right)} \\ & =\lim_{x\to -\infty }\limits\,\dfrac{-1-\dfrac{1}{x}}{\sqrt{1-\dfrac{1}{x}}+\sqrt{1+\dfrac{1}{x}}}=-\dfrac{1}{2} \\ \end{align}$