Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 5x} + x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {\sqrt {{x^2} + 5x} + x} \right)\left( {\sqrt {{x^2} + 5x} - x} \right)}}{{\sqrt {{x^2} + 5x} - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {{x^2} + 5x} \right) - {x^2}}}{{\sqrt {{x^2}\left( {1 + \frac{5}{x}} \right)} - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{5x}}{{\left| x \right|\sqrt {1 + \frac{5}{x}} - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{5x}}{{ - x\sqrt {1 + \frac{5}{x}} - x}}\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{5}{{ - \sqrt {1 + \frac{5}{x}} - 1}}\\
= \frac{5}{{ - \sqrt 1 - 1}} = - \frac{5}{2}\\
b,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {{x^2} + \sqrt {9{x^2} + 1} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } {x^2} + \mathop {\lim }\limits_{x \to + \infty } \sqrt {9{x^2} + 1} \\
= \left( { + \infty } \right) + \left( { + \infty } \right) = + \infty \\
c,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {x - \sqrt {{x^2} + x} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} - \left( {{x^2} + x} \right)}}{{x + \sqrt {{x^2} + x} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - x}}{{x + \sqrt {{x^2} + x} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 1}}{{1 + \sqrt {1 + \frac{1}{x}} }}\\
= \frac{{ - 1}}{{1 + \sqrt 1 }} = \frac{{ - 1}}{2}\\
d,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 5} - \sqrt {{x^2} + 3} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {{x^2} + 5} \right) - \left( {{x^2} + 3} \right)}}{{\sqrt {{x^2} + 5} + \sqrt {{x^2} + 3} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{2}{{\sqrt {{x^2} + 5} + \sqrt {{x^2} + 3} }}\\
= 0\\
\left( {\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 5} + \sqrt {{x^2} + 3} } \right) = + \infty } \right)\\
e,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt[3]{{{x^3} + x}} - x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {{x^3} + x} \right) - {x^3}}}{{{{\sqrt[3]{{{x^3} + x}}}^2} + \sqrt[3]{{{x^3} + x}}.x + {x^2}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{x}{{{{\sqrt[3]{{{x^3} + x}}}^2} + \sqrt[3]{{{x^3} + x}}.x + {x^2}}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\frac{1}{x}}}{{{{\sqrt[3]{{1 + \frac{1}{{{x^2}}}}}}^2} + \sqrt[3]{{1 + \frac{1}{{{x^2}}}}}.1 + 1}}\\
= \frac{0}{{1 + 1 + 1}} = 0
\end{array}\)
