Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 2x - 3}}{{x - 1}} = \frac{{{3^2} - 2.3 - 3}}{{3 - 1}} = \frac{0}{2} = 0\\
b,\\
\mathop {\lim }\limits_{x \to - 2} h\left( x \right) = \mathop {\lim }\limits_{x \to - 2} \frac{{2{x^3} + 15}}{{{{\left( {x + 2} \right)}^2}}}\\
\mathop {\lim }\limits_{x \to - 2} {\left( {x + 2} \right)^2} = {\left( { - 2 + 2} \right)^2} = 0\\
\mathop {\lim }\limits_{x \to - 2} \left( {2{x^3} + 15} \right) = \left( {2.{{\left( { - 2} \right)}^3} + 15} \right) = - 1\\
\Rightarrow \mathop {\lim }\limits_{x \to - 2} h\left( x \right) = \left( {\frac{{ - 1}}{0}} \right) = - \infty \\
c,\\
\mathop {\lim }\limits_{x \to - \infty } k\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \sqrt {4{x^2} - x + 1} = \mathop {\lim }\limits_{x \to - \infty } \left| x \right|.\sqrt {4 - \frac{1}{x} + \frac{1}{{{x^2}}}} = \left( { + \infty } \right).\sqrt 4 = + \infty \\
d,\\
\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \left( {{x^3} + {x^2} + 1} \right) = \mathop {\lim }\limits_{x \to - \infty } \left[ {{x^3}\left( {1 + \frac{1}{x} + \frac{1}{{{x^3}}}} \right)} \right] = \left( { - \infty } \right).1 = - \infty \\
e,\\
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} h\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{x - 15}}{{x + 2}}\\
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \left( {x - 15} \right) = \left( { - 2} \right) - 15 = - 17\\
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \left( {x + 2} \right) = {0^ + }\\
\Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ + }} \frac{{x - 15}}{{x + 2}} = \left( {\frac{{ - 17}}{{{0^ + }}}} \right) = - \infty \\
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} h\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \frac{{x - 15}}{{x + 2}}\\
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \left( {x - 15} \right) = \left( { - 2} \right) - 15 = - 17\\
\mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \left( {x + 2} \right) = {0^ - }\\
\Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 2} \right)}^ - }} \frac{{x - 15}}{{x + 2}} = \left( {\frac{{ - 17}}{{{0^ - }}}} \right) = + \infty
\end{array}\)
