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Q.limx0e2+xe2cos(exe2)x\displaystyle \lim_{x \to 0} \frac{e^{2+x} - e^2 \cos(e^x - e^2)}{x} equals

a
e2e^2
b
00
c
e2-e^2
d
2e22e^2

Correct Answer: Option A

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

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