Đáp án:
\[m = - 3\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{\sqrt {6 - x} - \sqrt[3]{{6 + x}}}}\\
= \mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{\left( {\sqrt {6 - x} - 2} \right) + \left( {2 - \sqrt[3]{{6 + x}}} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 2}}{{\frac{{{{\sqrt {6 - x} }^2} - {2^2}}}{{\sqrt {6 - x} + 2}} + \frac{{{2^3} - {{\sqrt[3]{{6 + x}}}^3}}}{{{2^2} + 2.\sqrt[3]{{6 + x}} + {{\sqrt[3]{{6 + x}}}^2}}}}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 2}}{{\frac{{6 - x - 4}}{{\sqrt {6 - x} + 2}} + \frac{{8 - 6 - x}}{{{{\sqrt[3]{{6 + x}}}^2} + 2\sqrt[3]{{6 + x}} + 4}}}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 2}}{{\frac{{2 - x}}{{\sqrt {6 - x} + 2}} + \frac{{2 - x}}{{{{\sqrt[3]{{6 + x}}}^2} + 2\sqrt[3]{{6 + x}} + 4}}}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 2}}{{\left( {2 - x} \right)\left( {\frac{1}{{\sqrt {6 - x} + 2}} + \frac{1}{{{{\sqrt[3]{{6 + x}}}^2} + 2\sqrt[3]{{6 + x}} + 4}}} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{ - 1}}{{\frac{1}{{\sqrt {6 - x} + 2}} + \frac{1}{{{{\sqrt[3]{{6 + x}}}^2} + 2\sqrt[3]{{6 + x}} + 4}}}}\\
= \dfrac{{ - 1}}{{\frac{1}{{\sqrt {6 - 2} + 2}} + \frac{1}{{{{\sqrt[3]{{6 + 2}}}^2} + 2.\sqrt[3]{{6 + 2}} + 4}}}}\\
= - 3\\
f\left( 2 \right) = m
\end{array}\)
Hàm số đã cho liên tục tại \(x = 2\) khi và chỉ khi:
\(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right) \Leftrightarrow m = - 3\)
Vậy \(m = - 3\)
