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