Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limitlimx→0+ (1−x)1x−e−1xa is equal to a nonzero real number, is

limx=0+ eln⁡(1−x)x−1exa =limx−0+ 1ee1+ln(1−x)x−1xa     =1elimx−0+ e1+ln(1−x)x−11+ln(1−x)x1+ln(1−x)xxa =1elimx−0+ 1+ln(1−x)xxa =1elimx−0+ ln(1−x)+xx(a+1) =1elimx−0+ −x−x22−x33−…….+xxa+1 Thus, a=1