Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
m,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {x + 2} - \sqrt {x - 2} } \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {\sqrt {x + 2} - \sqrt {x - 2} } \right)\left( {\sqrt {x + 2} + \sqrt {x - 2} } \right)}}{{\sqrt {x + 2} + \sqrt {x - 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{\left( {x + 2} \right) - \left( {x - 2} \right)}}{{\sqrt {x + 2} + \sqrt {x - 2} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{4}{{\sqrt {x + 2} + \sqrt {x - 2} }} = 0\\
\left( {\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {x + 2} + \sqrt {x - 2} } \right) = + \infty } \right)\\
n,\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} - 3x + 2} + x - 2} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( {\sqrt {{x^2} - 3x + 2} + \left( {x - 2} \right)} \right)\left( {\sqrt {{x^2} - 3x + 2} - \left( {x - 2} \right)} \right)}}{{\sqrt {{x^2} - 3x + 2} - \left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{{{\sqrt {{x^2} - 3x + 2} }^2} - {{\left( {x - 2} \right)}^2}}}{{\sqrt {{x^2}\left( {1 - \frac{3}{x} + \frac{2}{{{x^2}}}} \right)} - x + 2}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {{x^2} - 3x + 2} \right) - \left( {{x^2} - 4x + 4} \right)}}{{\left| x \right|.\sqrt {1 - \frac{3}{x} + \frac{2}{{{x^2}}}} - x + 2}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x - 2}}{{ - x.\sqrt {1 - \frac{3}{x} + \frac{2}{{{x^2}}}} - x + 2}}\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{1 - \frac{2}{x}}}{{ - \sqrt {1 - \frac{3}{x} + \frac{2}{{{x^2}}}} - 1 + \frac{2}{x}}}\\
= \frac{1}{{ - \sqrt 1 - 1}} = - \frac{1}{2}\\
o,\\
\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 2x}}{{3x + 1}} = \frac{{{2^2} - 2.2}}{{3.2 + 1}} = \frac{0}{7} = 0\\
p,\\
\mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2} - 3x + 3}}{{x - 2}}\\
\mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 3x + 3} \right) = {2^2} - 3.2 + 3 = 1\\
\left. \begin{array}{l}
\mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 2} \right) = 2 - 2 = 0\\
x \to {2^ - } \Rightarrow x < 0 \Rightarrow x - 2 < 0
\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 2} \right) = {0^ - }\\
\Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2} - 3x + 3}}{{x - 2}} = - \infty
\end{array}\)
