Giải thích các bước giải:
\(\begin{array}{l}
20,\\
\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} - 5x + 5} - 2x}}{{4{x^2} + 1}}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\sqrt {1 - \frac{5}{x} + \frac{5}{{{x^2}}}} - 2}}{{4x + \frac{1}{x}}}\\
= 0\\
\left( \begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {1 - \frac{5}{x} + \frac{5}{{{x^2}}}} - 2} \right) = \sqrt 1 - 2 = - 1\\
\mathop {\lim }\limits_{x \to + \infty } \left( {4x + \frac{1}{x}} \right) = + \infty
\end{array} \right)\\
23,\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {3{x^2} + x + 1} + x\sqrt 3 } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {\sqrt {3{x^2} + x + 1} + x\sqrt 3 } \right)\left( {\sqrt {3{x^2} + x + 1} - x\sqrt 3 } \right)}}{{\sqrt {3{x^2} + x + 1} - x\sqrt 3 }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3{x^2} + x + 1} \right) - {{\left( {x\sqrt 3 } \right)}^2}}}{{\sqrt {{x^2}\left( {3 + \frac{1}{x} + \frac{1}{{{x^2}}}} \right)} - x\sqrt 3 }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x + 1}}{{\left| x \right|\sqrt {3 + \frac{1}{x} + \frac{1}{{{x^2}}}} - x\sqrt 3 }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x + 1}}{{ - x\sqrt {3 + \frac{1}{x} + \frac{1}{{{x^2}}}} - x\sqrt 3 }}\,\,\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{1 + \frac{1}{x}}}{{ - \sqrt {3 + \frac{1}{x} + \frac{1}{{{x^2}}}} - \sqrt 3 }}\\
= \frac{1}{{ - 2\sqrt 3 }} = \frac{{ - \sqrt 3 }}{6}\\
29,\\
\mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{{x^3} + 2\sqrt 2 }}{{{x^2} - 2}}\\
= \mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{{x^3} + {{\left( {\sqrt 2 } \right)}^3}}}{{{x^2} - {{\left( {\sqrt 2 } \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{\left( {x + \sqrt 2 } \right)\left( {{x^2} - \sqrt 2 x + 2} \right)}}{{\left( {x - \sqrt 2 } \right)\left( {x + \sqrt 2 } \right)}}\\
= \mathop {\lim }\limits_{x \to - \sqrt 2 } \frac{{{x^2} - \sqrt 2 x + 2}}{{x - \sqrt 2 }}\\
= \frac{{{{\left( { - \sqrt 2 } \right)}^2} - \sqrt 2 .\left( { - \sqrt 2 } \right) + 2}}{{ - \sqrt 2 - \sqrt 2 }}\\
= \frac{6}{{ - 2\sqrt 2 }}\\
= - \frac{{3\sqrt 2 }}{2}
\end{array}\)
