Đáp án:
\[a = \frac{7}{{12}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3x + 1} - \sqrt[3]{{2x + 6}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {3x + 1} - 2} \right) + \left( {2 - \sqrt[3]{{2x + 6}}} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{3x + 1 - 4}}{{\sqrt {3x + 1} + 2}} + \frac{{8 - 2x - 6}}{{4 + 2\sqrt[3]{{2x + 6}} + {{\sqrt[3]{{2x + 6}}}^2}}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{3\left( {x - 1} \right)}}{{\sqrt {3x + 1} + 2}} - \frac{{2\left( {x - 1} \right)}}{{4 + 2\sqrt[3]{{2x + 6}} + {{\sqrt[3]{{2x + 6}}}^2}}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \left( {\frac{3}{{\sqrt {3x + 1} + 2}} - \frac{2}{{4 + 2\sqrt[3]{{2x + 6}} + {{\sqrt[3]{{2x + 6}}}^2}}}} \right)\\
= \frac{3}{{\sqrt {3.1} + 2}} - \frac{2}{{4 + 2.\sqrt[3]{{2.2 + 6}} + {{\sqrt[3]{{2.2 + 6}}}^2}}}\\
= \frac{3}{4} - \frac{2}{{4 + 4 + 4}} = \frac{7}{{12}}
\end{array}\)
Hàm số đã cho liên tục tại \({x_0} = 1\) khi và chỉ khi \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) \Leftrightarrow a = \frac{7}{{12}}\)
