Giải thích các bước giải:
\(\begin{array}{l}
b.\mathop {\lim }\limits_{x \to + \infty } \frac{{16{x^2} + 4x - 1 - 16{x^2} + 8x - 1}}{{\sqrt {16{x^2} + 4x - 1} + 4x - 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{12x - 2}}{{\sqrt {16{x^2} + 4x - 1} + 4x - 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{12 - \frac{2}{x}}}{{\sqrt {16 + \frac{4}{x} - \frac{1}{{{x^2}}}} + 4 - \frac{1}{x}}} = \frac{{12}}{{4 + 4}} = \frac{3}{2}\\
d.\mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} + 10x + 25 - {x^2} - 6x - 1}}{{x + 5 - \sqrt {{x^2} + 6x + 1} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{4x + 24}}{{x + 5 - \sqrt {{x^2} + 6x + 1} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{4 + \frac{{24}}{x}}}{{1 + \frac{5}{x} + \sqrt {1 + \frac{6}{x} + \frac{1}{{{x^2}}}} }}\\
= \frac{4}{{1 + 1}} = 2\\
e.\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} + 1 - {x^2} - 4x - 2}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} + 4x + 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 4x - 1}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} + 4x + 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 4 - \frac{1}{x}}}{{\sqrt {1 + \frac{1}{{{x^2}}}} + \sqrt {1 + \frac{4}{x} + \frac{2}{{{x^2}}}} }}\\
= \frac{{ - 4}}{{1 + 1}} = - 2\\
g.\mathop {\lim }\limits_{x \to - 1} \frac{{7{x^2} + 2 - 9{x^2}}}{{\left( {6 - x} \right)\left( {x + 1} \right)\left( {\sqrt {7{x^2} + 2} - 3x} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{2\left( {1 - x} \right)\left( {1 + x} \right)}}{{\left( {6 - x} \right)\left( {x + 1} \right)\left( {\sqrt {7{x^2} + 2} - 3x} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{2 - 2x}}{{\left( {6 - x} \right)\left( {\sqrt {7{x^2} + 2} - 3x} \right)}}\\
= \frac{{2 - 2.\left( { - 1} \right)}}{{\left( {6 + 1} \right)\left( {\sqrt {7 + 2} + 3} \right)}} = \frac{2}{{21}}\\
h.\mathop {\lim }\limits_{x \to 3} \frac{{5x + 1 - {x^2} - 2x - 1}}{{\left( {x - 3} \right)\left( {3x - 1} \right)\left( {\sqrt {5x + 1} + x + 1} \right)}}\\
\mathop { = \lim }\limits_{x \to 3} \frac{{x\left( {3 - x} \right)}}{{\left( {x - 3} \right)\left( {3x - 1} \right)\left( {\sqrt {5x + 1} + x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{ - x}}{{\left( {3x - 1} \right)\left( {\sqrt {5x + 1} + x + 1} \right)}}\\
= \frac{{ - 3}}{{\left( {3.3 - 1} \right)\left( {3 + 3 + 1} \right)}} = - \frac{3}{{80}}\\
k.\mathop {\lim }\limits_{x \to - 1} \frac{{\left( {x + 1} \right)\left( {x + 4} \right)\left( {3x - \sqrt {7{x^2} + 2} } \right)}}{{9{x^2} - 7{x^2} - 2}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {x + 1} \right)\left( {x + 4} \right)\left( {3x - \sqrt {7{x^2} + 2} } \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {x + 4} \right)\left( {3x - \sqrt {7{x^2} + 2} } \right)}}{{2x - 2}}\\
= \frac{{\left( { - 1 + 4} \right)\left( { - 3 - 3} \right)}}{{2.\left( { - 1} \right) - 2}} = \frac{9}{2}
\end{array}\)
