Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 110869 by Study last updated on 31/Aug/20

Answered by mathdave last updated on 31/Aug/20

$$letI=\underset{x\to 0}{\lim }x{!}^{\frac{1}{x}}$$$$bylogginbothsidewehave$$$$\ln I=\underset{x\to 0}{\lim }\frac{\ln x!}{x}butx!=\mathrm{\Gamma }(x+1)$$$$\ln I=\underset{x\to 0}{\lim }\frac{\ln \mathrm{\Gamma }(1+x)}{x}apply{lihospita}^{\prime }srule$$$$\ln I=\underset{x\to 0}{\lim }\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(1+x)}{\mathrm{\Gamma }(1+x)}but\mathrm{\Psi }(1+x)=\frac{{\mathrm{\Gamma }}^{{}^{\prime }}(1+x)}{\mathrm{\Gamma }(1+x)}$$$$\ln I=\underset{x\to 0}{\lim }\mathrm{\Psi }(1+x)=-\mathrm{\Psi }\left(1\right)$$$$I=e^{-\mathrm{\Psi }}$$$$\because \underset{x\to 0}{\lim }\sqrt[x]{x!}=e^{-\mathrm{\Psi }}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com