lim _{x rightarrow-infty} frac{3|x|-x}{|x|-2 | vedclass

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(B) Consider the first limit: $L_1 = \lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x}$.Since $x \rightarrow-\infty$,we have $|x| = -x$.Substituting

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$\frac{1}{3}$

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(B) Consider the first limit: $L_1 = \lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x}$.Since $x \rightarrow-\infty$,we have $|x| = -x$.Substituting this,we get $L_1 = \lim _{x \rightarrow-\infty} \frac{3(-x)-x}{(-x)-2 x} = \lim _{x \rightarrow-\infty} \frac{-4x}{-3x} = \frac{4}{3}$.Consider the second limit: $L_2 = \lim _{x \rightarrow 0} \frac{\log (1+x^3)}{\sin ^3 x}$.Using the standard limits $\lim _{u \rightarrow 0} \frac{\log (1+u)}{u} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:$L_2 = \lim _{x \rightarrow 0} \left( \frac{\log (1+x^3)}{x^3} \cdot \frac{x^3}{\sin ^3 x} \right) = 1 \cdot 1 = 1$.Now,subtracting the two results:$L_1 - L_2 = \frac{4}{3} - 1 = \frac{4-3}{3} = \frac{1}{3}$.

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

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