If $\lim\limits_{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=-1$ then

a = 1 a = 0 a = e none of these

a = 1

L’ Hospital Rule $\lim\limits_{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=\lim\limits_{x \rightarrow a} \frac{a^x \log a-a x^{a-1}}{x^x+x^x \log _e a}=-1$ $\frac{a^a \log a-a a^{a-1}}{a^a+a^a \log a}=\frac{\log a-1}{\log a+1}=-1$ It is satisfied only when a = 1. Hence (A) is the correct answer.