Giải thích các bước giải:
32,
Hai giới hạn đã cho là hai giới hạn hữu hạn nên ta có:
\(\begin{array}{l}
*)\\
\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} + ax + b}}{{{{\left( {x - 2} \right)}^2}}} = k\\
\Rightarrow {x^2} + ax + b = k{\left( {x - 2} \right)^2}\\
\Rightarrow {x^2} + ax + b = k\left( {{x^2} - 4x + 4} \right)\\
\Rightarrow \left\{ \begin{array}{l}
k = 1\\
a = - 4\\
b = 4
\end{array} \right.\\
*)\\
\mathop {\lim }\limits_{x \to - 2} \dfrac{{ - {x^2} + 2x + c}}{{4 - {x^2}}}\\
= \mathop {\lim }\limits_{x \to - 2} \dfrac{{ - x\left( {x + 2} \right) + 4\left( {x + 2} \right) + \left( {c - 8} \right)}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \dfrac{{\left( {x + 2} \right)\left( {4 - x} \right) + \left( {c - 8} \right)}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
= \mathop {\lim }\limits_{x \to - 2} \left( {\dfrac{{4 - x}}{{2 - x}} + \dfrac{{c - 8}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}} \right)\\
= \dfrac{{4 - \left( { - 2} \right)}}{{2 - \left( { - 2} \right)}} + \mathop {\lim }\limits_{x \to - 2} \dfrac{{c - 8}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
= \dfrac{3}{2} + \mathop {\lim }\limits_{x \to - 2} \dfrac{{c - 8}}{{\left( {2 - x} \right)\left( {2 + x} \right)}}\\
\Rightarrow c - 8 = 0 \Rightarrow c = 8\\
\Rightarrow a + b + c = 8\\
34,\\
\mathop {\lim }\limits_{x \to a} \dfrac{{{x^3} + 8}}{{x - a}} = 12\\
\mathop {\lim }\limits_{x \to a} \left( {x - a} \right) = 0\\
\Rightarrow \mathop {\lim }\limits_{x \to a} \left( {{x^3} + 8} \right) = 0 \Rightarrow {a^3} + 8 = 0 \Leftrightarrow a = - 2
\end{array}\)
