Đáp án:
a. lim=8
Giải thích các bước giải:
\(\begin{array}{l}
a.\mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)\left( {\sqrt {5x + 1} + 4} \right)}}{{5x + 1 - 16}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)\left( {\sqrt {5x + 1} + 4} \right)}}{{5x - 15}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)\left( {\sqrt {5x + 1} + 4} \right)}}{{5\left( {x - 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right)\left( {\sqrt {5x + 1} + 4} \right)\\
= \left( {3 - 2} \right)\left( {\sqrt {15 + 1} + 4} \right) = 8
\end{array}\)
b. Do hàm số liên tục tại x=1
\(\begin{array}{l}
\to \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = f\left( 1 \right)\\
Có:\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} {x^2} = {1^2} = 1\\
f\left( 1 \right) = - 1 + a\\
\mathop {Do:\lim }\limits_{x \to {1^ + }} f\left( x \right) = f\left( 1 \right)\\
\Leftrightarrow 1 = - 1 + a\\
\to a = 2
\end{array}\)
