The value of limx→0 ex2−cos⁡xx2 is equal to

limx→0 ex2−cos⁡xx2=limx→0 ex2−1+(1−cos⁡x)x2=limx→0 ex2−1+2sin2⁡x2x2=limx→0 ex2−1x2+limx→0 2sin⁡x2x2∵limx→0 ex−1x=1,limx→0 sin⁡kxx=k=1+2122=1+12=32Therefore, the correct answer is (3).