If a > 0 and Limx→aax−xaxx−aa=−1 then a =

Limit value is 0/0 ,Use L-Hospital rule ddx(ax)=axloga,ddx(xa)=a.xa−1ddx(xx)=xx(1+logx)limx→a axloga-axa-1xx(1+logx)-0=aaloga-aaaa(1+loga)=-1loga-1=-1-loga2loga=0a=1

If a > 0 and Limx→aax−xaxx−aa=−1 then a =

Limit value is 0/0 ,Use L-Hospital rule ddx(ax)=axloga,ddx(xa)=a.xa−1ddx(xx)=xx(1+logx)limx→a axloga-axa-1xx(1+logx)-0=aaloga-aaaa(1+loga)=-1loga-1=-1-loga2loga=0a=1

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If a > 0 and $\underset{x \to a}{L i m} \frac{a^{x} - x^{a}}{x^{x} - a^{a}} = - 1$ then a =

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answer is B.

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Detailed Solution

Limit value is 0/0 ,Use L-Hospital rule $\frac{d}{d x} (a^{x}) = a^{x} \log a, \frac{d}{d x} (x^{a}) = a . x^{a - 1}$ $\frac{d}{d x} (x^{x}) = x^{x} (1 + \log x)$ $\lim_{x \to a} \frac{a^{x} \log a - a x^{a - 1}}{x^{x} (1 + \log x) - 0} = \frac{a^{a} \log a - a^{a}}{a^{a} (1 + \log a)}$ =-1