Đáp án:
k) \( - \dfrac{1}{{12}}\)
Giải thích các bước giải:
\(\begin{array}{l}
k)\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {9 - 8 - x} \right)\left( {2x + \sqrt {5 - x} } \right)}}{{\left( {4{x^2} - 5x + x} \right)\left( {9 + \sqrt {8 + x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {1 - x} \right)\left( {2x + \sqrt {5 - x} } \right)}}{{\left( {4{x^2} - 4x} \right)\left( {9 + \sqrt {8 + x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {1 - x} \right)\left( {2x + \sqrt {5 - x} } \right)}}{{4x\left( {x - 1} \right)\left( {9 + \sqrt {8 + x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - \left( {2x + \sqrt {5 - x} } \right)}}{{4x\left( {9 + \sqrt {8 + x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - \left( {2.1 + \sqrt {5 - 1} } \right)}}{{4.1\left( {9 + \sqrt {8 + 1} } \right)}} = - \dfrac{1}{{12}}\\
l)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {{x^2} - x - 2} \right)\left( {\sqrt {4x + 1} + 3} \right)}}{{\left( {x + \sqrt {x + 2} } \right)\left( {4x + 1 - 9} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x + 1} \right)\left( {x - 2} \right)\left( {\sqrt {4x + 1} + 3} \right)}}{{4\left( {x - 2} \right)\left( {x + \sqrt {x + 2} } \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x + 1} \right)\left( {\sqrt {4x + 1} + 3} \right)}}{{4\left( {x + \sqrt {x + 2} } \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {2 + 1} \right)\left( {\sqrt {4.2 + 1} + 3} \right)}}{{4\left( {2 + \sqrt {2 + 2} } \right)}} = \dfrac{9}{8}\\
m)\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {{x^2} + 2x + 6 - {{\left( {4x - 1} \right)}^2}} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)\left( {\sqrt {{x^2} + 2x + 6} + 4x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 2x + 6 - 16{x^2} + 8x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)\left( {\sqrt {{x^2} + 2x + 6} + 4x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 15{x^2} + 10x + 5}}{{\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)\left( {\sqrt {{x^2} + 2x + 6} + 4x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{5\left( {1 - x} \right)\left( {3x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)\left( {\sqrt {{x^2} + 2x + 6} + 4x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 5\left( {3x + 1} \right)}}{{\left( {{x^2} + x - 1} \right)\left( {\sqrt {{x^2} + 2x + 6} + 4x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 5\left( {3 + 1} \right)}}{{\left( {1 + 1 - 1} \right)\left( {\sqrt {1 + 2 + 6} + 4 - 1} \right)}} = - \dfrac{{10}}{3}
\end{array}\)
