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Question Number 115664 by bemath last updated on 27/Sep/20

$$If\underset{x\to p}{\lim }\frac{p^x-x^p}{x^x-p^p}=1thenp=?$$

Answered by Olaf last updated on 27/Sep/20

$$x=p+u$$$$\underset{x\to p}{\lim }\frac{e^{x\ln p}-e^{p\ln x}}{e^{x\ln x}-e^{p\ln p}}$$$$\underset{x\to p}{\lim }\frac{e^{\frac{x\ln p+p\ln x}{2}}(e^{\frac{x\ln p-p\ln x}{2}}-e^{-\frac{x\ln p-p\ln x}{2}})}{e^{\frac{x\ln x+p\ln p}{2}}(e^{\frac{x\ln x-p\ln p}{2}}-e^{-\frac{x\ln x-p\ln p}{2}})}$$$$\underset{x\to p}{\lim }{e}^{\frac{x\ln \frac{p}{x}+p\ln \frac{x}{p}}{2}}\left[\frac{\sinh \left(\frac{x\ln p-p\ln x}{2}\right)}{\sinh \left(\frac{x\ln x-p\ln p}{2}\right)}\right]$$$$\underset{x\to p}{\lim }1\times \left[\frac{\sinh \left(\frac{x\ln p-p\ln x}{2}\right)}{\sinh \left(\frac{x\ln x-p\ln p}{2}\right)}\right]$$$$\underset{x\to p}{\lim }\left[\frac{\left(\frac{\ln p-\frac{p}{x}}{2}\right)\cosh \left(\frac{x\ln p-p\ln x}{2}\right)}{\left(\frac{\ln x+1}{2}\right)\cosh \left(\frac{x\ln x-p\ln p}{2}\right)}\right]$$$$=\frac{\ln p-1}{\ln p+1}=\frac{\ln \left(\frac{p}{e}\right)}{\ln \left(ep\right)}$$$$\frac{\ln \left(\frac{p}{e}\right)}{\ln \left(ep\right)}=1\Rightarrow \frac{p}{e}=ep?$$$${\mathbf{I}}^{\prime }\mathbf{m}\mathbf{surely}\mathbf{wrong}.$$$$$$$$$$$$$$

Answered by Dwaipayan Shikari last updated on 27/Sep/20

$$\underset{x\to \mathrm{p}}{\lim }\frac{{\mathrm{p}}^{\mathrm{x}}-{\mathrm{x}}^{\mathrm{p}}}{{\mathrm{x}}^{\mathrm{x}}-{\mathrm{p}}^{\mathrm{p}}}=\frac{{\mathrm{p}}^{\mathrm{x}}\mathrm{logp}-{\mathrm{px}}^{\mathrm{p}-1}}{{\mathrm{x}}^{\mathrm{x}}(1+\mathrm{logx})}=\frac{{\mathrm{p}}^{\mathrm{p}}(\mathrm{logp}-1)}{{\mathrm{p}}^{\mathrm{p}}(\mathrm{logp}+1)}=1$$$$\frac{\mathrm{logp}-1}{\mathrm{logp}+1}=1$$$$\mathrm{Oh}\mathrm{i}\mathrm{think}\mathrm{it}\mathrm{is}\mathrm{independent}\mathrm{of}\mathrm{P}!$$

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