Bài tập 4.24 trang 165 SBT Toán 11

a) \(\mathop {\lim }\limits_{x \to  - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} - 3x} }}{{x + 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {1 - \frac{3}{x}} }}{{x\left( {1 + \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {1 - \frac{3}{x}} }}{{1 + \frac{2}{x}}} =  - 1\)

\(\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {{x^2} - 3x} }}{{x + 2}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\sqrt {1 - \frac{3}{x}} }}{{x\left( {1 + \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {1 - \frac{3}{x}} }}{{1 + \frac{2}{x}}} = 1\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \left( {x + \sqrt {{x^2} - x + 1} } \right) = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x - 1}}{{x - \sqrt {{x^2} - x + 1} }}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {1 - \frac{1}{x}} \right)}}{{x\left( {1 + \sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}} } \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{1 - \frac{1}{x}}}{{1 + \sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}} }} = \frac{1}{2}
\end{array}\)

\(\mathop {\lim }\limits_{x \to  + \infty } \left( {x + \sqrt {{x^2} - x + 1} } \right) = \mathop {\lim }\limits_{x \to  + \infty } x\left( {1 + \sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}} } \right) =  + \infty \)

c)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} - x}  - \sqrt {{x^2} + 1} } \right) = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x - 1}}{{\sqrt {{x^2} - x}  + \sqrt {{x^2} + 1} }}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( { - 1 - \frac{1}{x}} \right)}}{{ - x\left( {\sqrt {1 - \frac{1}{x}}  + \sqrt {1 + \frac{1}{x}} } \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - 1 - \frac{1}{x}}}{{ - \left( {\sqrt {1 - \frac{1}{x}}  + \sqrt {1 + \frac{1}{x}} } \right)}} = \frac{1}{2}
\end{array}\)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} - x}  - \sqrt {{x^2} + 1} } \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - x - 1}}{{\sqrt {{x^2} - x}  + \sqrt {{x^2} + 1} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( { - 1 - \frac{1}{x}} \right)}}{{x\left( {\sqrt {1 - \frac{1}{x}}  + \sqrt {1 + \frac{1}{x}} } \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 1 - \frac{1}{x}}}{{\sqrt {1 - \frac{1}{x}}  + \sqrt {1 + \frac{1}{x}} }} =  - \frac{1}{2}
\end{array}\)

-- Mod Toán 11 HỌC247