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Q.limx0loge(1+4sin2x)[x]\lim_{x \to 0} \dfrac{\log_e (1 + 4\sin^2 x)}{[x]} (where [][\cdot] is GIF) equals

a
00
b
22
c
44
d
\infty

Correct Answer: Option C

The correct solution involves applying the fundamental concept to derive the final value step by step...

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