Đáp án:
\[\mathop {\lim }\limits_{x \to {3^ + }} \sqrt {{x^2} - 9} .\frac{{2x + 1}}{{x - 3}} = + \infty \]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ + }} \sqrt {{x^2} - 9} .\frac{{2x + 1}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} .\frac{{2x + 1}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \sqrt {x - 3} .\sqrt {x + 3} .\frac{{2x + 1}}{{x - 3}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\sqrt {x + 3} .\left( {2x + 1} \right)}}{{\sqrt {x - 3} }}\\
\mathop {\lim }\limits_{x \to {3^ + }} \left( {\sqrt {x + 3} \left( {2x + 1} \right)} \right) = \sqrt {3 + 3} .\left( {2.3 + 1} \right) = 7\sqrt 6 \\
\left. \begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ + }} \sqrt {x - 3} = \sqrt {3 - 3} = 0\\
x \to {3^ + } \Rightarrow x - 3 > 0 \Rightarrow \sqrt {x - 3} > 0
\end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \sqrt {x - 3} = {0^ + }\\
\Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\sqrt {x + 3} .\left( {2x + 1} \right)}}{{\sqrt {x - 3} }} = + \infty \\
\Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \sqrt {{x^2} - 9} .\frac{{2x + 1}}{{x - 3}} = + \infty
\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to {3^ + }} \sqrt {{x^2} - 9} .\frac{{2x + 1}}{{x - 3}} = + \infty \)
