Limit _ x 3 (x^ 3 +27 ) (x-2) (x^ 2 -9 ) =
Mathematics · JEE Advanced · NTA Exams — Limit, Continuity and Differentiability
\[\operatorname{Limit}_{x \rightarrow 3} \frac{\left(x^{3}+27\right) \ln (x-2)}{\left(x^{2}-9\right)}=\]
- – 8
- 8
- 9
- – 9
Answer
(C) 9
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