Question

${\lim }_{x\to a}\frac{a^x-x^a}{x^x-a^a}=-1$, then

• A. a=1
• B. a=0
• C. a=e
• D. none of these

Answer 1

Answer: (A)

${\lim }_{x\to a}\frac{a^x-x^a}{x^x-a^a}={\lim }_{x\to a}\frac{a^x\log a-a{x}^{a-1}}{x^x\left(1+\log x\right)}$ (L Hospital Rule) $=\frac{a^a\left(\log -1\right)}{a^a\left(\log a+1\right)}=\frac{\log a-1}{\log a+1}=-1$ (give) $\Rightarrow \log a-1=-\log a-1$ $\Rightarrow 2\log a=0\Rightarrow a=1$