_ x 0 (27 + x) ^ _ 1 3 - 3 9 - (27 + x) ^ 2 3 equals. - Vidyadip

MCQ

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

A
$-\frac {1}{3}$
B
$\frac {1}{6}$
✓
$-\frac {1}{6}$
D
$\frac {1}{3}$

✓

Answer

Correct option: C.

$-\frac {1}{6}$

c
Let $L = \,\,\,\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {27 + x} \right)}^{\frac{1}{3}}} - 3}}{{9 - {{\left( {27 + x} \right)}^{\frac{2}{3}}}}}$

Here $'L'$ is in the indeterminate from i.e.,$\frac{0}{0}$

$\therefore $ usinh the $L'$ Hosoital rule we get:

$L = \,\,\,\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{3}{{\left( {27 + x} \right)}^{\frac{{ - 2}}{3}}}}}{{ - \frac{2}{3}{{\left( {27 + x} \right)}^{\frac{{ - 1}}{3}}}}} = \frac{{\frac{1}{3} \times {{\left( {27} \right)}^{\frac{{ - 2}}{3}}}}}{{\frac{{ - 2}}{3} \times {{\left( {27} \right)}^{\frac{{ - 1}}{3}}}}} = - \frac{1}{6}$

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