JEE Mains · Maths · STD 11 - 12. limits
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{(27 + x)}^{_{\frac{1}{3}}}} - 3}}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}\) equals.
- A \(-\frac {1}{3}\)
- B \(\frac {1}{6}\)
- C \(-\frac {1}{6}\)
- D \(\frac {1}{3}\)
Answer & Solution
Correct Answer
(C) \(-\frac {1}{6}\)
Step-by-step Solution
Detailed explanation
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:…
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