If-lim-x-p-p-x-x-p-x-x-p-p-1-then-p-

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

I f \lim_{x \to p} \frac{p^{x} - x^{p}}{x^{x} - p^{p}} = 1 t h e n p = ?

Answered by Olaf last updated on 27/Sep/20

x = p + u

\lim_{x \to p} \frac{e^{x \ln p} - e^{p \ln x}}{e^{x \ln x} - e^{p \ln p}}

\lim_{x \to p} \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}})}

\lim_{x \to p} e^{\frac{x \ln \frac{p}{x} + p \ln \frac{x}{p}}{2}} [\frac{\sinh (\frac{x \ln p - p \ln x}{2})}{\sinh (\frac{x \ln x - p \ln p}{2})}]

\lim_{x \to p} 1 \times [\frac{\sinh (\frac{x \ln p - p \ln x}{2})}{\sinh (\frac{x \ln x - p \ln p}{2})}]

\lim_{x \to p} [\frac{(\frac{\ln p - \frac{p}{x}}{2}) \cosh (\frac{x \ln p - p \ln x}{2})}{(\frac{\ln x + 1}{2}) \cosh (\frac{x \ln x - p \ln p}{2})}]

= \frac{\ln p - 1}{\ln p + 1} = \frac{\ln (\frac{p}{e})}{\ln (e p)}

\frac{\ln (\frac{p}{e})}{\ln (e p)} = 1 \Rightarrow \frac{p}{e} = e p ?

I^{'} m surely wrong .

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

\lim_{x \to p} \frac{p^{x} - x^{p}}{x^{x} - p^{p}} = \frac{p^{x} logp - {px}^{p - 1}}{x^{x} (1 + logx)} = \frac{p^{p} (logp - 1)}{p^{p} (logp + 1)} = 1

\frac{logp - 1}{logp + 1} = 1

Oh i think it is independent of P!