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Question Number 112455 by john santu last updated on 08/Sep/20

$$\left(1\right)\underset{x\to \mathrm{p}}{\lim }\frac{{\mathrm{p}}^{\mathrm{x}}-{\mathrm{x}}^{\mathrm{p}}}{{\mathrm{x}}^{\mathrm{x}}-{\mathrm{p}}^{\mathrm{p}}}?$$$$\left(2\right)Thereare4identicalmathbooks,$$$$2identicalphysicsbooksand2identical$$$$chemistrybooks.Howmanywaysto$$$$compiletheeightbooksonthe$$$$conditionofthesamebooksarenot$$$$mutuallyadjacent?$$

Answered by bobhans last updated on 08/Sep/20

$$\mathrm{let}\mathrm{y}={\mathrm{x}}^{\mathrm{x}}\Rightarrow \ln \mathrm{y}=\mathrm{xln}\mathrm{x};\to \frac{{\mathrm{y}}^{\prime }}{\mathrm{y}}=\ln \mathrm{x}+1$$$${\mathrm{y}}^{\prime }={\mathrm{x}}^{\mathrm{x}}(1+\ln \mathrm{x})$$$$\left(1\right)\underset{x\to \mathrm{p}}{\lim }\frac{{\mathrm{p}}^{\mathrm{x}}\ln \left(\mathrm{p}\right)-{\mathrm{px}}^{\mathrm{p}-1}}{{\mathrm{x}}^{\mathrm{x}}(1+\ln \mathrm{x})}=\frac{{\mathrm{p}}^{\mathrm{p}}\ln \mathrm{p}-\mathrm{p}.{\mathrm{p}}^{\mathrm{p}-1}}{{\mathrm{p}}^{\mathrm{p}}(1+\ln \mathrm{p})}=\frac{{\mathrm{p}}^{\mathrm{p}}(\ln \mathrm{p}-1)}{{\mathrm{p}}^{\mathrm{p}}(1+\ln \mathrm{p})}$$$$=\frac{\ln \mathrm{p}-1}{\ln \mathrm{p}+1}=\frac{\ln \left(\frac{\mathrm{p}}{\mathrm{e}}\right)}{\ln \left(\mathrm{pe}\right)}=\log {}_{\mathrm{pe}}\left(\frac{\mathrm{p}}{\mathrm{e}}\right)$$

Answered by mr W last updated on 08/Sep/20

$$\left(2\right)$$$$2\times \frac{4!}{2!2!}+C_1^3\times 2!\times 2!=24$$

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