Đáp án:$=\infty$
Giải thích các bước giải:
$lim_{x\rightarrow 0}\dfrac{1-tan6x}{3x}=lim_{x\rightarrow 0}(\dfrac{1}{3x}-\dfrac{tan6x}{3x})$
$=lim_{x\rightarrow 0}(\dfrac{1}{3x}-\dfrac{\frac{sin6x}{cos6x}}{3x})$
$=lim_{x\rightarrow 0}(\dfrac{1}{3x}-\dfrac{\frac{2sin3xcos3x}{cos6x}}{3x})$
$=lim_{x\rightarrow 0}(\dfrac{1}{3x}-\dfrac{\frac{sin3x}{3x}.2cos3x}{cos6x}$
$=\infty-\dfrac{1.2.1}{1}$
$=\infty$
