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Question-122350




Question Number 122350 by shaker last updated on 16/Nov/20
Answered by liberty last updated on 16/Nov/20
  lim_(x→a)  ((x^a −a^x )/(a^x −a^a )) = lim_(x→a)  ((ax^(a−1) −a^x .ln a)/(a^x .ln a))   = ((a.a^(a−1) −a^a .ln a)/(a^a .ln a)) = ((a^a (1−ln a))/(a^a .ln a))   = ((1−ln a)/(ln a)) = log _e (a)−1.▲
$$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{x}^{{a}} −{a}^{{x}} }{{a}^{{x}} −{a}^{{a}} }\:=\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{ax}^{{a}−\mathrm{1}} −{a}^{{x}} .\mathrm{ln}\:{a}}{{a}^{{x}} .\mathrm{ln}\:{a}} \\ $$$$\:=\:\frac{{a}.{a}^{{a}−\mathrm{1}} −{a}^{{a}} .\mathrm{ln}\:{a}}{{a}^{{a}} .\mathrm{ln}\:{a}}\:=\:\frac{{a}^{{a}} \left(\mathrm{1}−\mathrm{ln}\:{a}\right)}{{a}^{{a}} .\mathrm{ln}\:{a}} \\ $$$$\:=\:\frac{\mathrm{1}−\mathrm{ln}\:{a}}{\mathrm{ln}\:{a}}\:=\:\mathrm{log}\:_{{e}} \left({a}\right)−\mathrm{1}.\blacktriangle \\ $$

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