Đáp án:
a. \(\frac{1}{2}\)
b. 5
c. -3
d. -1/9
Giải thích các bước giải:
\(\begin{array}{l}
a.\mathop {\lim }\limits_{x \to 4} \frac{{x\left( {x - 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \frac{x}{{x + 4}}\\
= \frac{4}{8} = \frac{1}{2}\\
b.\mathop {\lim }\limits_{x \to 0} \frac{{x\left( {5 + 9x} \right)}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \left( {5 + 9x} \right) = 5 + 9.0 = 5\\
c.\mathop {\lim }\limits_{x \to - 2} \frac{{\left( {2x - 5} \right)\left( {x + 2} \right)}}{{3\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{2x - 5}}{3} = \frac{{ - 4 - 5}}{3} = - 3\\
d.\mathop {\lim }\limits_{x \to 4} \frac{{\left( {3 - x} \right)\left( {x - 4} \right)}}{{\left( {x - 4} \right)\left( {x + 5} \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \frac{{3 - x}}{{x + 5}} = \frac{{ - 1}}{9}\\
e.\mathop {\lim }\limits_{x \to - 1} \frac{{5\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {4{x^2} - 3x + 6} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{5x - 5}}{{4{x^2} - 3x + 6}} = \frac{{ - 10}}{{13}}\\
f.\mathop {\lim }\limits_{x \to 5} \frac{{6x - 5 - 25}}{{\left( {x - 5} \right)\left( {2x + 3} \right)\left( {\sqrt {6x - 5} + 5} \right)}}\\
= \mathop {\lim }\limits_{x \to 5} \frac{{6\left( {x - 5} \right)}}{{\left( {x - 5} \right)\left( {2x + 3} \right)\left( {\sqrt {6x - 5} + 5} \right)}}\\
= \mathop {\lim }\limits_{x \to 5} \frac{6}{{\left( {2x + 3} \right)\left( {\sqrt {6x - 5} + 5} \right)}}\\
= \frac{6}{{130}} = \frac{3}{{65}}\\
g.\mathop {\lim }\limits_{x \to - 3} \frac{{x\left( {x + 3} \right)\left( {4 + \sqrt {7 - 3x} } \right)}}{{16 - 7 + 3x}}\\
= \mathop {\lim }\limits_{x \to - 3} \frac{{x\left( {x + 3} \right)\left( {4 + \sqrt {7 - 3x} } \right)}}{{3\left( {x + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to - 3} \frac{{x\left( {4 + \sqrt {7 - 3x} } \right)}}{3}\\
= \frac{{ - 24}}{3} = - 8
\end{array}\)
