Answer:
Option B
Explanation:
$\lim_{x \rightarrow -\infty}\frac{3|x|-x}{|x|-2x}-\lim_{x \rightarrow 0}\frac{\log (1+x^{3})}{\sin^{3}x}$
= $\lim_{x \rightarrow -\infty}\frac{-3x-x}{-x-2x}-\lim_{x \rightarrow 0}\frac{\frac{\log (1+x^{3})}{x^{3}}\times x^{3}}{\left(\frac{\sin x}{x}\right)^{3}\times x^{3}}$
[ $\because$ |x|= -x, x<0]
= $\frac{3+1}{1+2}-1= \frac{4}{3}-1=\frac{1}{3}$
