limx→a xa−axxx−aa is equal to
1+logea1−logea
loge(e/a)loge(ae)
logc(a/e)loge(ae)
none of these
We have,
limx→a xa−axxx−aa 00 form
=limx→a axa−1−axlogaxx(1+logx)−0 [Using L' Hospital's Rule]
=aa−aalogaaa(1+loga)=1−loga1+loga=loge(e/a)loge(ae)