Limit[(x^a - a^x)/(x^x - a^a), x -> a] Anyone knows how to solve this? I applied La'Hospitals rule but didnt really proceed far.
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Originally Posted by champrock Limit[(x^a - a^x)/(x^x - a^a), x -> a] Anyone knows how to solve this? I applied La'Hospitals rule but didnt really proceed far. It should have worked as long as a > 0, applying L'Hopital's rule once gives $\displaystyle \lim_{x \to a} \frac{ax^{a-1} - a^x \ln a}{x^x (\ln x+1)} = \frac{1 - \ln a}{1 + \ln a}$
Did you do it correctly? $\displaystyle \lim _{x \to a} \frac{{x^a - a^x }} {{x^x - a^a }}\mathop = \limits^H \lim _{x \to a} \frac{{ax^{a - 1} - a^x \ln (a)}}{{x^x \left( {\ln (x) + 1} \right)}}$
OK, really silly mistake! i did uptil that point correctly but didnt put in the values of X after that! was somehow expecting something simpler. hehe. thanks coming for u guys!
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