Limit _ x 0 (4^ x -1 )^ 3 ( x p ) (1+ x^ 2 3 ) =
Mathematics · JEE Advanced · NTA Exams — Limit, Continuity and Differentiability
\[\operatorname{Limit}_{x \rightarrow 0} \frac{\left(4^{x}-1\right)^{3}}{\sin \left(\frac{x}{p}\right) \ln \left(1+\frac{x^{2}}{3}\right)}=\]
- 9 p (log 4)
- 3 p (log 4)3
- 12 p (log 4)3
- 27 p (log 4)2
Answer
(B) 3 p ( l og 4) 3
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