Bài tập 31 trang 159 SGK Toán 11 NC

a)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \sqrt 2 } \frac{{{x^3} + 2\sqrt 2 }}{{{x^2} - 2}} = \mathop {\lim }\limits_{x \to  - \sqrt 2 } \frac{{{x^3} + {{\left( {\sqrt 2 } \right)}^3}}}{{{x^2} - {{\left( {\sqrt 2 } \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to  - \sqrt 2 } \frac{{\left( {x + \sqrt 2 } \right)\left( {{x^2} - x\sqrt 2  + 2} \right)}}{{\left( {x + \sqrt 2 } \right)\left( {x - \sqrt 2 } \right)}}\\
 = \mathop {\lim }\limits_{x \to  - \sqrt 2 } \frac{{{x^2} - x\sqrt 2  + 2}}{{x - \sqrt 2 }} = \frac{{ - 3\sqrt 2 }}{2}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 27x}}{{2{x^2} - 3x - 9}}\\
 = \mathop {\lim }\limits_{x \to 3} \frac{{x\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)}}{{\left( {x - 3} \right)\left( {2x + 3} \right)}}
\end{array}\\
{ = \mathop {\lim }\limits_{x \to 3} \frac{{x\left( {{x^2} + 3x + 9} \right)}}{{2x + 3}} = 9}
\end{array}\)

c)

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to  - 2} \frac{{{x^4} - 16}}{{{x^2} + 6x + 8}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {{x^2} + 4} \right)\left( {{x^2} - 4} \right)}}{{\left( {x + 2} \right)\left( {x + 4} \right)}}\\
 = \mathop {\lim }\limits_{x \to  - 2} \frac{{\left( {{x^2} + 4} \right)\left( {x - 2} \right)}}{{x + 4}} =  - 16
\end{array}\)

d)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x}  + x - 1}}{{\sqrt {{x^2} - {x^3}} }}\\
 = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt {1 - x}  - \left( {1 - x} \right)}}{{\left| x \right|\sqrt {1 - x} }}
\end{array}\\
{ = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{1 - \sqrt {1 - x} }}{{\left| x \right|}} = 1}
\end{array}\)

-- Mod Toán 11 HỌC247