The value of limx→0 (1+x)1/x−ex, is

We have,(1+x)1/x=elog⁡(1+x)x=e1xx−x22+x33−…⇒ (1+x)1/x=e1−x2+x23−⋯=e⋅e−x2+x23⋯limx→0 (1+x)1/x−ex=limx→0 e⋅e−x2+x23⋯⋅−ex=elimx→0 e−x/2+x2/3+…−1−x2+x23+…×−x2+x2x+…x=−12 e