Question

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

(a) 1

(b) 0

(c) $e$

(d) $\left(\frac{1}{e}\right)$

Answer 1

SOLUTION — $\lim _{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=\lim _{x \rightarrow a} \frac{a^x \log _e a-a x^{a-1}}{x^x\left(1+\log _e x\right)}$(by L'Hospital's rule)

$\begin{array}{lc}\Rightarrow & -1=\frac{a^a \log _e a-a^a}{a^a\left(\log _e a+1\right)}=\frac{\log _e a-1}{\log _e a+1} \\\therefore & \log _e a-1=-\log _e a-1 \\\Rightarrow & 2 \log _e a=0 \\\Rightarrow & a=e^0 \Rightarrow a=1\end{array}$

So, The correct option of this question will be (A).