Giải thích các bước giải:
\(\begin{array}{l}
1,\\
\mathop {\lim }\limits_{x \to {{\left( { - 4} \right)}^ + }} \frac{{2x + 1}}{{x + 4}}\\
\mathop {\lim }\limits_{x \to {{\left( { - 4} \right)}^ + }} \left( {2x + 1} \right) = 2.\left( { - 4} \right) + 1 = - 7\\
\mathop {\lim }\limits_{x \to {{\left( { - 4} \right)}^ + }} \left( {x + 4} \right) = {0^ + }\\
\Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 4} \right)}^ + }} \frac{{2x + 1}}{{x + 4}} = \left( {\frac{{ - 7}}{{{0^ + }}}} \right) = - \infty \\
2,\\
\mathop {\lim }\limits_{x \to 5} \frac{{{x^2} - 5x}}{{{x^2} - 25}} = \mathop {\lim }\limits_{x \to 5} \frac{{x\left( {x - 5} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \mathop {\lim }\limits_{x \to 5} \frac{x}{{x + 5}} = \frac{5}{{5 + 5}} = \frac{1}{2}\\
3,\\
\mathop {\lim }\limits_{x \to 2} \frac{{2 - \sqrt {x + 2} }}{{{x^2} - 3x + 2}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{{2^2} - \left( {x + 2} \right)}}{{2 + \sqrt {x + 2} }}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{\frac{{ - \left( {x - 2} \right)}}{{2 + \sqrt {x + 2} }}}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{\left( {x - 1} \right)\left( {2 + \sqrt {x + 2} } \right)}}\\
= \frac{{ - 1}}{{\left( {2 - 1} \right).\left( {2 + \sqrt {2 + 2} } \right)}}\\
= \frac{{ - 1}}{{1.4}}\\
= - \frac{1}{4}
\end{array}\)
