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Q.

The value of limx→0 (1−cos⁡2x)sin⁡5xx2sin⁡3x is

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a

103

b

310

c

65

d

56

answer is A.

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Detailed Solution

limx→0 (1−cos⁡2x)sin⁡5xx2sin⁡3x=limx→0 2sin2⁡xsin⁡5xx2sin⁡3x=limx→0 2sin2⁡xx2sin⁡5xxsin⁡3xx=limx→0 2sin⁡xx2×5limx→0 sin⁡5x5x3limx→0 sin⁡3x3x=2×53=103 ∵limx→0 sin⁡xx=1
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The value of limx→0 (1−cos⁡2x)sin⁡5xx2sin⁡3x is