Question

If $ \displaystyle\lim_{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a} = -1 $ then $ a $ is equal to

• A. 0
• B. e
• C. 2
• D. None of the above

Answer 1

Answer: (D)

We have, $\displaystyle\lim _{x \rightarrow a} \frac{a^{x}-x^{a}}{x^{x}-a^{a}}=-1$
$\Rightarrow \displaystyle\lim _{x \rightarrow a} \frac{a^{x} \log a-a x^{a-1}}{x^{x}(\log x+1)}=-1$
[using L'Hospital's rule]
$\Rightarrow \frac{a^{a} \log a-a \cdot a^{a-1}}{a^{a}(\log a+1)}=-1$
$\Rightarrow \frac{\log a-1}{\log a+1}=-1$
$\Rightarrow \log a-1=-\log a-1$
$\Rightarrow 2 \log a=0$
$\Rightarrow \log a=0$
$\Rightarrow a=e^{0}$
$\Rightarrow a=1$