Question Number 126801 by john_santu last updated on 24/Dec/20
$$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\:\frac{{x}^{{a}} −{a}^{{x}} }{{a}^{{x}} −{a}^{{a}} }\:=\:\:;\:{a}>\mathrm{0}\: \\ $$$$\: \\ $$
Answered by liberty last updated on 24/Dec/20
$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left({x}^{{a}} −{a}^{{x}} \right)}{\frac{{d}}{{dx}}\left({a}^{{x}} −{a}^{{a}} \right)}\:=\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{ax}^{{a}−\mathrm{1}} −\mathrm{ln}\:{a}.\left({a}^{{x}} \right)}{\mathrm{ln}\:{a}.\left({a}^{{x}} \right)} \\ $$$$\:=\:\frac{{a}.{a}^{{a}−\mathrm{1}} −{a}^{{a}} .\mathrm{ln}\:{a}}{{a}^{{a}} .\mathrm{ln}\:{a}}\:=\:\frac{{a}^{{a}} \left(\mathrm{1}−\mathrm{ln}\:{a}\right)}{{a}^{{a}} .\mathrm{ln}\:{a}}=\:\frac{\mathrm{1}−\mathrm{ln}\:{a}}{\mathrm{ln}\:{a}} \\ $$