lim_{x rightarrow -infty} frac{3|x|-x}{|x|-2x | vedclass

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(A) Let $L = \lim_{x \rightarrow -\infty} \frac{3|x|-x}{|x|-2x} - \lim_{x \rightarrow 0} \frac{\log(1+x^3)}{\sin^3 x}$.For the first limit,as $x \righ

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$\frac{1}{3}$

Vedclass · Answer

(A) Let $L = \lim_{x ightarrow -\infty} \frac{3|x|-x}{|x|-2x} - \lim_{x ightarrow 0} \frac{\log(1+x^3)}{\sin^3 x}$.For the first limit,as $x ightarrow -\infty$,$|x| = -x$. Thus,$\lim_{x ightarrow -\infty} \frac{3(-x)-x}{-x-2x} = \lim_{x ightarrow -\infty} \frac{-4x}{-3x} = \frac{4}{3}$.For the second limit,we use the standard limits $\lim_{x ightarrow 0} \frac{\log(1+x^3)}{x^3} = 1$ and $\lim_{x ightarrow 0} \frac{\sin x}{x} = 1$.So,$\lim_{x ightarrow 0} \frac{\log(1+x^3)}{\sin^3 x} = \lim_{x ightarrow 0} \left( \frac{\log(1+x^3)}{x^3} 	imes \frac{x^3}{\sin^3 x} ight) = 1 	imes 1 = 1$.Therefore,$L = \frac{4}{3} - 1 = \frac{1}{3}$.

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

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