mathop {lim }limits_{x to 0} frac{{{{(27 + x) | vedclass

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(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{(27 + x)}^{\frac{1}{3}}} - 3}}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$Substituting $x = 0$,we get th

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$-\frac{1}{6}$

Vedclass · Answer

(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{{{(27 + x)}^{\frac{1}{3}}} - 3}}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$Substituting $x = 0$,we get the indeterminate form $\frac{0}{0}$.Applying $L$'Hospital's rule by differentiating the numerator and denominator with respect to $x$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}({{(27 + x)}^{\frac{1}{3}}} - 3)}}{{\frac{d}{{dx}}(9 - {{(27 + x)}^{\frac{2}{3}}})}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{3}{{(27 + x)}^{ - \frac{2}{3}}}}}{{ - \frac{2}{3}{{(27 + x)}^{ - \frac{1}{3}}}}}$$L = \mathop {\lim }\limits_{x 	o 0} -\frac{1}{2} {(27 + x)}^{ - \frac{2}{3} + \frac{1}{3}}$$L = -\frac{1}{2} {(27)}^{ - \frac{1}{3}} = -\frac{1}{2} 	imes \frac{1}{3} = -\frac{1}{6}$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(27 + x)}^{\frac{1}{3}}}} - 3}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$ equals.

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