Bài tập 59 trang 178 SGK Toán 11 NC

Bài tập 59 trang 178 SGK Toán 11 NC

a)

\(\mathop {\lim }\limits_{x \to  - 2} \sqrt[3]{{\frac{{2{x^4} + 3x + 1}}{{{x^2} - x + 2}}}} = \sqrt[3]{{\frac{{2.{{\left( { - 2} \right)}^4} + 3.\left( { - 2} \right) + 1}}{{{{\left( { - 2} \right)}^2} - \left( { - 2} \right) + 2}}}} = \frac{3}{2}\)

b)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} - x + 5} }}{{2x - 1}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left| x \right|\sqrt {1 - \frac{1}{x} + \frac{5}{{{x^2}}}} }}{{x\left( {2 - \frac{1}{x}} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {1 - \frac{1}{x} + \frac{5}{{{x^2}}}} }}{{2 - \frac{1}{x}}} =  - \frac{1}{2}
\end{array}\)

c) Với x < - 3, ta có: 

\(\frac{{{x^4} + 1}}{{{x^2} + 4x + 3}} = \frac{{{x^4} + 1}}{{x + 1}}.\frac{1}{{x + 3}}\)

Vì \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \frac{{{x^4} + 1}}{{x + 1}} = \frac{{82}}{{ - 2}} =  - 41 < 0,\)

\(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \frac{1}{{x + 3}} =  - \infty \) nên \(\mathop {\lim }\limits_{x \to {{\left( { - 3} \right)}^ - }} \frac{{{x^4} + 1}}{{{x^2} + 4x + 3}} =  + \infty \)

d)

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

\(\mathop {\lim }\limits_{x \to 2} \sqrt {\frac{{x + 4}}{{4 - x}}}  = \sqrt {\frac{6}{2}}  = \sqrt 3  > 0\) nên \(\mathop {\lim }\limits_{x \to 2} \frac{3}{{{{\left( {x - 2} \right)}^2}}}\sqrt {\frac{{x + 4}}{{4 - x}}}  =  + \infty \)

e)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{\sqrt {8 + 2x}  - 2}}{{\sqrt {x + 2} }}\\
 = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{8 + 2x - 4}}{{\sqrt {x + 2} \left( {\sqrt {8 + 2x}  + 2} \right)}}
\end{array}\\
\begin{array}{l}
 = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{2\left( {x + 2} \right)}}{{\sqrt {x + 2} \left( {\sqrt {8 + 2x}  + 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{2\sqrt {x + 2} }}{{\sqrt {8 + 2x}  + 2}} = \frac{0}{4} = 0
\end{array}
\end{array}\)

f)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} + x}  - \sqrt {4 + {x^2}} } \right) = \mathop {\lim }\limits_{x \to  - \infty } \frac{{{x^2} + x - 4 - {x^2}}}{{\sqrt {{x^2} + x}  + \sqrt {4 + {x^2}} }}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x - 4}}{{\left| x \right|\sqrt {1 + \frac{1}{x}}  + \left| x \right|\sqrt {\frac{4}{{{x^2}}} + 1} }}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {1 - \frac{4}{x}} \right)}}{{ - x\left( {\sqrt {1 + \frac{1}{x}}  + \sqrt {\frac{4}{{{x^2}}} + 1} } \right)}}\\
 =  - \mathop {\lim }\limits_{x \to  - \infty } \frac{{1 - \frac{4}{x}}}{{\sqrt {1 + \frac{1}{x}}  + \sqrt {1 + \frac{4}{{{x^2}}}} }} =  - \frac{1}{2}
\end{array}\)

-- Mod Toán 11