limx→a xa−axxx−aa is equal to

We have,limx→a xa−axxx−aa                   00 form=limx→a axa−1−axlog⁡axx(1+log⁡x)−0                       [Using L' Hospital's Rule]=aa−aalog⁡aaa(1+log⁡a)=1−log⁡a1+log⁡a=loge⁡(e/a)loge⁡(ae)