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Question-136741




Question Number 136741 by JulioCesar last updated on 25/Mar/21
Answered by Dwaipayan Shikari last updated on 25/Mar/21
lim_(x→0) ((x!−1)/x)=((Γ(x+1)−1)/x)=((Γ′(x+1))/1)=Γ′(1)=−γ
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}!−\mathrm{1}}{{x}}=\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{1}}{{x}}=\frac{\Gamma'\left({x}+\mathrm{1}\right)}{\mathrm{1}}=\Gamma'\left(\mathrm{1}\right)=−\gamma \\ $$

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