Let x = 1 + δ, where δ is very small and tends to zero as x tends to 1.
Then x1/m = (1 + δ)1/m → 1 + δ/m as δ → 0
Simnilarly x1/n → 1 + δ/n
This means (x1/m -1)/(x1/n - 1) → (1 + δ/m - 1)/(1 + δ/n - 1) → (δ/m)/(δ/n) → n/m
So the limit of (x1/m -1)/(x1/n - 1) as x → 1 is n/m
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