Question

If $\underset{x\to a}{lim}\frac{a^x-x^a}{x^x-a^a}=-1$, then $a=?$

• A. $1$
• B. $0$
• C. $e$
• D. $\frac{1}{e}$

Answer 1

The correct option is A

$1$

Explanation for the correct option.

Step 1: Apply Limit Hospital Rule.

By L Hospital rule $\underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$.

By differentiating $\frac{a^x-x^a}{x^x-a^a}$ with respect to $x$, we get

$\frac{a^x\log a-a{x}^{a-1}}{x^x\left(1+\log x\right)-0}$

Now,

$\underset{x\to a}{lim}\frac{a^x\log a-a{x}^{a-1}}{x^x\left(1+\log x\right)}=-1......\left(1\right)$

Step 2: Find the value of $a$.

By applying limit in $\left(1\right)$, we get

$\begin{array}{rcl}\frac{a^a\log a-a·a^{a-1}}{a^a\left(1+\log a\right)}& =& -1\\ & \Rightarrow & \frac{a^a\log a-a^{a-1+1}}{a^a\left(1+\log a\right)}=-1\\ \Rightarrow \frac{a^a\log a-a^a}{a^a\left(1+\log a\right)}& =& -1\\ \Rightarrow \frac{a^a\left(\log a-1\right)}{a^a\left(1+\log a\right)}& =& -1\\ \Rightarrow \log a-1& =& -\log a-1\\ \Rightarrow 2\log a& =& 0\\ \Rightarrow \log a& =& 0\\ \Rightarrow a& =& 1\end{array}$

Hence, option A is correct.