Question

$\underset{x\to a}{lim}\frac{x^a-a^x}{x^x-a^a}$ is equal to

• A. $\frac{1+{\log }_{e}a}{1-{\log }_{e}a}$
• B. $\frac{{\log }_e\left(e/a\right)}{{\log }_{e}ae}$
• C. $\frac{{\log }_e\left(a/e\right)}{{\log }_e\left(ae\right)}$
• D. none of these

Answer 1

Answer: (B)

$\frac{{\log }_e\left(e/a\right)}{{\log }_{e}ae}$

We have,
$\underset{x\to a}{lim}\frac{x^a-a^x}{x^x-a^a}$ $\left[\frac{0}{0}\right]$
$=\underset{x\to a}{lim}\frac{a{x}^{a-1}-a^x\log a}{x^x(1+\log x)-0}$ [Using L' Hospital's Rule]
$=\frac{a^a-a^a\log a}{a^a(1+\log a)}=\frac{1-\log a}{1+\log a}=\frac{lo{g}_e\left(e/a\right)}{{\log }_e\left(ae\right)}$
Hence, option 'B' is correct.