Giải thích các bước giải:
\(\begin{array}{l}
2,\\
\mathop {\lim }\limits_{x \to {{\left( { - 5} \right)}^ - }} \left( {x - 5} \right) = \left( { - 5} \right) - 5 = - 10 < 0\\
\mathop {\lim }\limits_{x \to {{\left( { - 5} \right)}^ - }} \left( {x + 5} \right) = {0^ - }\\
\Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 5} \right)}^ - }} \frac{{x - 5}}{{x + 5}} = + \infty \\
3,\\
\mathop {\lim }\limits_{x \to + \infty } \frac{{x - 3}}{{{x^2} - 9}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{x + 3}} = 0\\
\left( {\mathop {\lim }\limits_{x \to + \infty } \left( {x + 3} \right) = + \infty } \right)\\
4,\\
\mathop {\lim }\limits_{x \to {1^ + }} \left( {x + 3} \right) = 1 + 3 = 4 > 0\\
\mathop {\lim }\limits_{x \to {1^ + }} \left( {x - 1} \right) = {0^ + }\\
\Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x + 3}}{{x - 1}} = + \infty \\
5,\\
\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {4x - 3} \right) = 4.2 - 3 = 5\\
\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 2x + 3} \right) = {2^2} - 2.2 + 3 = 3\\
\Rightarrow 2.\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) + 3.\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 2.5 + 3.3 = 19
\end{array}\)
