Giải thích các bước giải:
\(\begin{array}{l}
4,\\
\mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} + x - 6}}{{{x^2} - 4}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x + 3} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \frac{{x + 3}}{{x + 2}}\\
\mathop {\lim }\limits_{x \to - 2} \left( {x + 3} \right) = - 2 + 3 = 1\\
\mathop {\lim }\limits_{x \to - 2} \left( {x + 2} \right) = - 2 + 2 = 0\\
\Rightarrow \mathop {\lim }\limits_{x \to - 2} \frac{{x + 3}}{{x + 2}} = \left( {\frac{1}{0}} \right) = \infty \\
5,\\
\mathop {\lim }\limits_{x \to {1^ - }} \frac{{5x + 1}}{{x - 1}}\\
\mathop {\lim }\limits_{x \to {1^ - }} \left( {5x + 1} \right) = 5.1 + 1 = 6\\
\mathop {\lim }\limits_{x \to {1^ - }} \left( {x - 1} \right) = {1^ - } - 1 = {0^ - }\\
\Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} \frac{{5x + 1}}{{x - 1}} = \left( {\frac{6}{{{0^ - }}}} \right) = - \infty \\
6,\\
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + 2x} - 1}}{{2x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{1 + 2x - 1}}{{\sqrt {1 + 2x} + 1}}}}{{2x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{2x}}{{\sqrt {1 + 2x} + 1}}}}{{2x}}\\
= \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {1 + 2x} + 1}}\\
= \frac{1}{{\sqrt {1 + 2.0} + 1}}\\
= \frac{1}{2}\\
7,\\
\mathop {\lim }\limits_{x \to {2^ + }} \frac{{5x - 3}}{{2 - x}}\\
\mathop {\lim }\limits_{x \to {2^ + }} \left( {5x - 3} \right) = 5.2 - 3 = 7\\
\mathop {\lim }\limits_{x \to {2^ + }} \left( {2 - x} \right) = 2 - {2^ + } = {0^ - }\\
\Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} \frac{{5x - 3}}{{2 - x}} = \left( {\frac{7}{{{0^ - }}}} \right) = - \infty
\end{array}\)
