#1**+5 **

Let x = 1 + δ, where δ is very small and tends to zero as x tends to 1.

Then x^{1/m} = (1 + δ)^{1/m} → 1 + δ/m as δ → 0

Simnilarly x^{1/n} → 1 + δ/n

This means (x^{1/m} -1)/(x^{1/n} - 1) → (1 + δ/m - 1)/(1 + δ/n - 1) → (δ/m)/(δ/n) → n/m

So the limit of (x^{1/m} -1)/(x^{1/n} - 1) as x → 1 is n/m

.

Alan
Apr 2, 2015

#1**+5 **

Best Answer

Let x = 1 + δ, where δ is very small and tends to zero as x tends to 1.

Then x^{1/m} = (1 + δ)^{1/m} → 1 + δ/m as δ → 0

Simnilarly x^{1/n} → 1 + δ/n

This means (x^{1/m} -1)/(x^{1/n} - 1) → (1 + δ/m - 1)/(1 + δ/n - 1) → (δ/m)/(δ/n) → n/m

So the limit of (x^{1/m} -1)/(x^{1/n} - 1) as x → 1 is n/m

.

Alan
Apr 2, 2015