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Question-158847




Question Number 158847 by mathlove last updated on 09/Nov/21
Answered by mr W last updated on 09/Nov/21
lim_(x→a) ((x^x −a^a )/(x−a))=a^a (ln a+1)  lim_(y→x) ((y^y −x^x )/(y−x))=x^x (ln x+1)
$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{x}} −{a}^{{a}} }{{x}−{a}}={a}^{{a}} \left(\mathrm{ln}\:{a}+\mathrm{1}\right) \\ $$$$\underset{{y}\rightarrow{x}} {\mathrm{lim}}\frac{{y}^{{y}} −{x}^{{x}} }{{y}−{x}}={x}^{{x}} \left(\mathrm{ln}\:{x}+\mathrm{1}\right) \\ $$
Commented by mathlove last updated on 09/Nov/21
solution?
$${solution}? \\ $$
Commented by mr W last updated on 09/Nov/21
lim_(x→a) ((x^x −a^a )/(x−a))=lim_(x→a) ((x^x (ln x+1))/1)=a^a (ln a+1)
$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{x}} −{a}^{{a}} }{{x}−{a}}=\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{x}} \left(\mathrm{ln}\:{x}+\mathrm{1}\right)}{\mathrm{1}}={a}^{{a}} \left(\mathrm{ln}\:{a}+\mathrm{1}\right) \\ $$

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