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Q.

limx→1 (2−x)Tanπx2

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a

e1π

b

e2π

c

−e2π

d

e

answer is B.

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Detailed Solution

limx→1 (2−x)tan⁡πx2 1∞ form =elimx→1 tan⁡πx2[2−x−1]=elimx→1 tan⁡πx2[1−x]=elimx→1 1−xcot⁡πx2=elimx→1-1-cosec2⁡πx2π2      L hospital rule=e−1−12π=e2π
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