Đáp án:
\[\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 3}}{{2 - \sqrt {3x - 5} }} = - \frac{4}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 3}}{{2 - \sqrt {3x - 5} }}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left( {x - 3} \right)\left( {2 + \sqrt {3x - 5} } \right)}}{{\left( {2 - \sqrt {3x - 5} } \right)\left( {2 + \sqrt {3x - 5} } \right)}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left( {x - 3} \right)\left( {2 + \sqrt {3x - 5} } \right)}}{{{2^2} - \left( {3x - 5} \right)}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left( {x - 3} \right)\left( {2 + \sqrt {3x - 5} } \right)}}{{9 - 3x}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left( {x - 3} \right)\left( {2 + \sqrt {3x - 5} } \right)}}{{3\left( {3 - x} \right)}}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{{2 + \sqrt {3x - 5} }}{{ - 3}}\\
= \frac{{2 + \sqrt {3.3 - 5} }}{{ - 3}} = - \frac{4}{3}
\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 3}}{{2 - \sqrt {3x - 5} }} = - \frac{4}{3}\)
