Question

$\underset{y\to x}{lim}\frac{y^y-x^x}{y-x}=$

• A. $x{y}^{x-1}$
• B. $x^x(1+\log x)$
• C. $y.x^{y-1}$
• D. $xy$

Answer 1

The correct option is B $x^x(1+\log x)$

$\underset{y\to x}{lim}\frac{y^y-x^x}{y-x}$
It is of the form $\frac{0}{0}$
Let $u=y^y$
$\Rightarrow \log u=y\log y$
$\frac{1}{u}\frac{du}{dy}=1+\log y$
$\frac{du}{dy}=y^y(1+\log y)$
Applying L-Hospital's rule,
$=\underset{y\to x}{lim}\frac{y^y(1+\log y)}{1}=x^x(1+\log x)$