Let p = limx→0+ (1+tan2√x)1/2x then log p is equal to: 2, 1, 1/2, 1/4

p = limx→0+ (1+tan2√x)1/2x, 1∞ form
= limx→0+ (1+tan2√x)1/tan2√x × tan2√x⁄2x
= {limx→0+ (1+tan2√x)1/tan2√x} × {limx→0+ tan2√x/2x}
= elimx→0+ 1/2 × (tan√x/√x)2 = e1/2 × 12 = e1/2
Therefore ln p = ln e1/2 = 1/2