Giải thích các bước giải:
\(\begin{array}{l}
1,\\
\mathop {\lim }\limits_{x \to - 2} \frac{{{x^3} + 5{x^2} - 12}}{{{x^2} + 4x + 4}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {{x^3} + 2{x^2}} \right) + \left( {3{x^2} - 12} \right)}}{{{{\left( {x + 2} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2}\left( {x + 2} \right) + 3\left( {x - 2} \right)\left( {x + 2} \right)}}{{{{\left( {x + 2} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x + 2} \right).\left( {{x^2} + 3x - 6} \right)}}{{{{\left( {x + 2} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} + 3x - 6}}{{x + 2}}\\
\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} + 3x - 6} \right) = {\left( { - 2} \right)^2} + 3.\left( { - 2} \right) - 6 = - 8\\
\mathop {\lim }\limits_{x \to - 2} \left( {x + 2} \right) = 0\\
\Rightarrow \mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} + 3x - 6}}{{x + 2}} = \infty \\
2,\\
\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}\\
= \mathop {\lim }\limits_{x \to 7} \frac{{\frac{{{2^2} - \left( {x - 3} \right)}}{{2 + \sqrt {x - 3} }}}}{{\left( {x - 7} \right)\left( {x + 7} \right)}}\\
= \mathop {\lim }\limits_{x \to 7} \frac{{\frac{{7 - x}}{{2 + \sqrt {x - 3} }}}}{{\left( {x - 7} \right)\left( {x + 7} \right)}}\\
= \mathop {\lim }\limits_{x \to 7} \frac{{ - 1}}{{\left( {2 + \sqrt {x - 3} } \right)\left( {x + 7} \right)}}\\
= \frac{{ - 1}}{{\left( {2 + \sqrt {7 - 3} } \right)\left( {7 + 7} \right)}}\\
= \frac{{ - 1}}{{4.14}}\\
= \frac{{ - 1}}{{56}}\\
3,\\
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1} - \sqrt {{x^2} + x + 1} }}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {x + 1} - 1} \right) + \left( {1 - \sqrt {{x^2} + x + 1} } \right)}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{x + 1 - 1}}{{\sqrt {x + 1} + 1}} + \frac{{1 - {x^2} - x - 1}}{{1 + \sqrt {{x^2} + x + 1} }}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{x}{{\sqrt {x + 1} + 1}} - \frac{{x\left( {x + 1} \right)}}{{1 + \sqrt {{x^2} + x + 1} }}}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{{\sqrt {x + 1} + 1}} - \frac{{x + 1}}{{1 + \sqrt {{x^2} + x + 1} }}} \right]\\
= \frac{1}{{\sqrt {0 + 1} + 1}} - \frac{{0 + 1}}{{1 + \sqrt {{0^2} + 0 + 1} }}\\
= \frac{1}{2} - \frac{1}{2}\\
= 0
\end{array}\)
