Limit of higher number Limx → 1 (x ^ m-1) / (x ^ n-1) limit, the answer is m / N, how to do without the law of Robita

The simplest is
Write x ^ m and x ^ n respectively to refer to the situation
(e^(mlnx)-1)/(e^(nlnx)-1)
This can be replaced by infinitesimal, and so it becomes
mlnx/nlnx
Finally, eliminate LNX
We get m / n

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