Giải thích các bước giải:
\(\begin{array}{l}
a.\mathop {\lim }\limits_{x \to + \infty } \frac{{3x + 1}}{{\sqrt {{x^2} + 3x + 1} + x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{3 + \frac{1}{x}}}{{\sqrt {1 + \frac{3}{x} + \frac{1}{{{x^2}}}} + 1}}\\
= \frac{3}{2}\\
b.\mathop {\lim }\limits_{x \to + \infty } \frac{{2x - 1 + 4x - 1}}{{\sqrt {4{x^2} + 2x - 1} + 2x - 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{6 - \frac{2}{x}}}{{\sqrt {4 + \frac{2}{x} - \frac{1}{{{x^2}}}} + 2 - \frac{1}{x}}}\\
= \frac{3}{2}\\
c.\mathop {\lim }\limits_{x \to - \infty } \frac{{2x + 1}}{{\sqrt {9{x^2} + 2x + 1} - 3x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{2 + \frac{1}{x}}}{{ - \sqrt {9 + \frac{2}{x} + \frac{1}{{{x^2}}}} - 3}}\\
= - \frac{1}{3}\\
e.\mathop {\lim }\limits_{x \to + \infty } \frac{{1 - 4x - 2}}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} + 4x + 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 4 - \frac{1}{x}}}{{\sqrt {1 + \frac{1}{{{x^2}}}} + \sqrt {1 + \frac{4}{x} + \frac{2}{{{x^2}}}} }} = - 2\\
f.\mathop {\lim }\limits_{x \to - \infty } \frac{{2 - 1}}{{\sqrt {4{x^2} + 2} - \sqrt {4{x^2} + 1} }} = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{{\sqrt {4{x^2} + 2} - \sqrt {4{x^2} + 1} }}\\
= 0\\
g.\mathop {\lim }\limits_{x \to 1} \frac{{5\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\left( {\sqrt {5{x^2} + 4} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{5x + 5}}{{\left( {{x^2} + x + 1} \right)\left( {\sqrt {5{x^2} + 4} + 3} \right)}} = \frac{{10}}{{3\left( {3 + 3} \right)}} = \frac{5}{9}
\end{array}\)
