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

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(D) Given the limit: $\lim _{x \rightarrow-\infty} \frac{3|x|^3-x^2+2|x|-5}{-5|x|^3+3 x^2-2|x|+7}$Since $x \rightarrow -\infty$,we have $|x| = -x$. Le

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$\frac{-3}{5}$

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(D) Given the limit: $\lim _{x \rightarrow-\infty} \frac{3|x|^3-x^2+2|x|-5}{-5|x|^3+3 x^2-2|x|+7}$Since $x \rightarrow -\infty$,we have $|x| = -x$. Let $x = -t$,where $t \rightarrow \infty$. Then $|x| = t$.Substituting these into the expression:$= \lim _{t \rightarrow \infty} \frac{3t^3 - (-t)^2 + 2t - 5}{-5t^3 + 3(-t)^2 - 2t + 7}$$= \lim _{t \rightarrow \infty} \frac{3t^3 - t^2 + 2t - 5}{-5t^3 + 3t^2 - 2t + 7}$Divide the numerator and denominator by $t^3$:$= \lim _{t}$ ${\rightarrow \infty} \frac{3 - \frac{1}{t} + \frac{2}{t^2} - \frac{5}{t^3}}{-5 + \frac{3}{t} - \frac{2}{t^2} + \frac{7}{t^3}}$As $t \rightarrow \infty$,all terms with $t$ in the denominator approach $0$.$= \frac{3 - 0 + 0 - 0}{-5 + 0 - 0 + 0} = -\frac{3}{5}$

$\lim _{x \rightarrow-\infty} \frac{3|x|^3-x^2+2|x|-5}{-5|x|^3+3 x^2-2|x|+7} = $

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