Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to 2} \frac{{x + 3}}{{{x^2} + x + 4}} = \frac{{2 + 3}}{{{2^2} + 2 + 4}} = \frac{5}{{10}} = \frac{1}{2}\\
b,\\
\mathop {\lim }\limits_{x \to - 3} \frac{{{x^2} + 5x + 6}}{{{x^2} + 3x}} = \mathop {\lim }\limits_{x \to - 3} \frac{{\left( {x + 2} \right)\left( {x + 3} \right)}}{{x\left( {x + 3} \right)}} = \mathop {\lim }\limits_{x \to - 3} \frac{{x + 2}}{x} = \frac{{ - 3 + 2}}{{ - 3}} = \frac{1}{3}\\
c,\\
\mathop {\lim }\limits_{x \to + \infty } \left( { - {x^3} + {x^2} - 2x + 1} \right) = \mathop {\lim }\limits_{x \to + \infty } \left[ {{x^3}.\left( { - 1 + \frac{1}{x} - \frac{2}{{{x^2}}} + \frac{1}{{{x^3}}}} \right)} \right]\\
\mathop {\lim }\limits_{x \to + \infty } {x^3} = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \left( { - 1 + \frac{1}{x} - \frac{2}{{{x^2}}} + \frac{1}{{{x^3}}}} \right) = - 1\\
\Rightarrow \mathop {\lim }\limits_{x \to + \infty } \left( { - {x^3} + {x^2} - 2x + 1} \right) = - \infty \\
d,\\
\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} - 2x + 4} - x}}{{3x - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|.\sqrt {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} - x}}{{3x - 1}}\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{\left( { - x} \right).\sqrt {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} - x}}{{3x - 1}}\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} - 1}}{{3 - \frac{1}{x}}}\\
= \frac{{ - \sqrt 1 - 1}}{3} = - \frac{2}{3}
\end{array}\)
