Cho các giới hạn \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = 2\); \(\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right) = 3\), khi đó \(\mathop {\lim }\limits_{x \to {x_0}} \left[ {3f\left( x \right) - 4g\left( x \right)} \right]\) bằng

Chọn B

Ta có:

\[\begin{array}{l}S = 9 + 3 + 1 + \frac{1}{3} + \frac{1}{9} + \cdots + \frac{1}{{{3^{n - 3}}}} + \cdots \\ = 9(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + \frac{1}{{81}} + ... + \frac{1}{{{3^n} - 1}}) = 9.\frac{1}{{1 - \frac{1}{3}}} = \frac{{27}}{2}\end{array}\]

  Chọn A

Ta có

 \[\begin{array}{l}\sin \left( {x + \frac{\pi }{3}} \right) + \sin 2x = 0 \Leftrightarrow \sin \left( {x + \frac{\pi }{3}} \right) = - \sin 2x = \sin \left( { - 2x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{3} = - 2x + k2\pi \\x + \frac{\pi }{3} = \pi + 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{9} + \frac{{k2\pi }}{3}\\x = \frac{{ - 2\pi }}{3} - k2\pi \end{array} \right.\\\end{array}\]

Vì nghiệm thuộc khoảng \[\left( {0;\,2\pi } \right)\] nên chọn \(k = 0,k = 1,k = 2,k = 3\)