Question

$\displaystyle \lim_{x \to y}$ $\left(\frac{x^{y}-y^{x}}{x^{x}-y^{y}}\right)$, is equal to

• A. $\frac{1+log\,y}{log\,y}$
• B. $\frac{1-log\,y}{1+log\,y}$
• C. $\frac{1-log\,y}{log\left(ey\right)}$
• D. $log (ex)$

Answer 1

Answer: (C)

$\displaystyle \lim_{x \to y}$ $\left(\frac{x^{y}-y^{x}}{x^{x}-y^{y}}\right)$ $(\frac{0}{0}$ form$)$
Using $L’$ Hospital Rule, we get
$\displaystyle \lim_{x \to y}$ $\frac{yx^{y-1}-y^{x}\,log\,y}{x^{x}\,log\left(ex\right)}$
$=\frac{1-log\,y}{log\left(ey\right)}$