Prove-that-lim-0-1-x-ln-x-

Question Number 10564 by FilupS last updated on 18/Feb/17

Prove that :

\lim_{ϵ \to 0} \frac{- 1 + x^{ϵ}}{ϵ} = \ln (x)

Answered by lee last updated on 21/Feb/17

\lim_{ε \to 0} \frac{x^{ε} - 1}{ε - 0} = \frac{\partial f}{\partial ε} (x, 0), f (x, ε) = x^{ε}

\frac{d}{d ε} x^{ε} = \frac{d}{d ε} e^{ε \ln x} = \frac{d}{d ε} (ε \ln x) e^{ε \ln x}

= x^{ε} \ln x

∴ ε \to 0, \ln x

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