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

Prove that: lim_(ε→0) ((−1+x^ε )/ε) = ln(x)

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\epsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{1}+{x}^{\epsilon} }{\epsilon}\:=\:\mathrm{ln}\left({x}\right) \\ $$

Answered by lee last updated on 21/Feb/17

lim_(ε→0) ((x^ε −1)/(ε−0))=(∂f/∂ε)(x,0),f(x,ε)=x^ε (d/dε)x^ε =(d/dε)e^(εln x) =(d/dε)(εln x)e^(εln x) =x^ε ln x ∴ε→0, ln x

$$ \\ $$$$ \\ $$$$\underset{\varepsilon\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\varepsilon} −\mathrm{1}}{\varepsilon−\mathrm{0}}=\frac{\partial{f}}{\partial\varepsilon}\left({x},\mathrm{0}\right),{f}\left({x},\varepsilon\right)={x}^{\varepsilon} \\ $$$$\frac{{d}}{{d}\varepsilon}{x}^{\varepsilon} =\frac{{d}}{{d}\varepsilon}{e}^{\varepsilon\mathrm{ln}\:{x}} =\frac{{d}}{{d}\varepsilon}\left(\varepsilon\mathrm{ln}\:{x}\right){e}^{\varepsilon\mathrm{ln}\:{x}} \\ $$$$={x}^{\varepsilon} \mathrm{ln}\:{x} \\ $$$$\therefore\varepsilon\rightarrow\mathrm{0},\:\mathrm{ln}\:{x} \\ $$

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