Discussion Forum

$\lim_{x \rightarrow -\infty}\frac{3|x|-x}{|x|-2x}-\lim_{x \rightarrow 0}\frac{\log (1+x^{3})}{\sin^{3}x}=$

Answer:

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]

 = $\frac{3+1}{1+2}-1= \frac{4}{3}-1=\frac{1}{3}$