Giải thích các bước giải:
$\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 0 - } \dfrac{{\sqrt {{x^2} - {x^3}} }}{{2x}}\\
= \mathop {\lim }\limits_{x \to 0 - } \dfrac{{\left| x \right|\sqrt {1 - x} }}{{2x}}\\
= \mathop {\lim }\limits_{x \to 0 - } \dfrac{{ - x\sqrt {1 - x} }}{{2x}}\\
= \mathop {\lim }\limits_{x \to 0 - } \dfrac{{ - \sqrt {1 - x} }}{2}\\
= \dfrac{{ - 1}}{2}\\
2)\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} + 3}}{{{x^3} - 2x + 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\dfrac{2}{x} + \dfrac{3}{{{x^3}}}}}{{1 - \dfrac{2}{{{x^2}}} + \dfrac{1}{{{x^3}}}}}\\
= \dfrac{{2.0 + 3.0}}{{1 - 2.0 + 0}}\\
= 0\\
3)\mathop {\lim }\limits_{x \to 2} \dfrac{{x - \sqrt {3x - 2} }}{{{x^2} - 4}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 3x + 2}}{{\left( {x + \sqrt {3x - 2} } \right)\left( {{x^2} - 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x + \sqrt {3x - 2} } \right)\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 1}}{{\left( {x + \sqrt {3x - 2} } \right)\left( {x + 2} \right)}}\\
= \dfrac{{2 - 1}}{{\left( {2 + \sqrt {3.2 - 2} } \right)\left( {2 + 2} \right)}}\\
= \dfrac{1}{{16}}\\
4)\mathop {\lim }\limits_{x \to - \infty } x\left( {\sqrt {{x^2} + 5} + x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } x\left( { - x\sqrt {1 + \dfrac{5}{{{x^2}}}} + x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } {x^2}\left( {1 - \sqrt {1 + \dfrac{5}{{{x^2}}}} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } {x^2}.\dfrac{{1 - \left( {1 + \dfrac{5}{{{x^2}}}} \right)}}{{1 + \sqrt {1 + \dfrac{5}{{{x^2}}}} }}\\
= \mathop {\lim }\limits_{x \to - \infty } {x^2}.\dfrac{{\dfrac{{ - 5}}{{{x^2}}}}}{{1 + \sqrt {1 + \dfrac{5}{{{x^2}}}} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 5}}{{1 + \sqrt {1 + \dfrac{5}{{{x^2}}}} }}\\
= \dfrac{{ - 5}}{{1 + \sqrt {1 + 5.0} }}\\
= \dfrac{{ - 5}}{2}\\
5)\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 1} - \sqrt {{x^2} + x + 1} }}{x}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{x + 1 - \left( {{x^2} + x + 1} \right)}}{{x\left( {\sqrt {x + 1} + \sqrt {{x^2} + x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - {x^2}}}{{x\left( {\sqrt {x + 1} + \sqrt {{x^2} + x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - x}}{{\sqrt {x + 1} + \sqrt {{x^2} + x + 1} }}\\
= \dfrac{{ - 0}}{{\sqrt {0 + 1} + \sqrt {{0^2} + 0 + 1} }}\\
= 0
\end{array}$
