limy→x yy−xxy−x=

Use L-hospital ruleddyyy=yy(1+log⁡y) and ddyxx=0limy→x yy−xxy−x=limy→xyy(1+log⁡y)-01-0=xx(1+logx)

limy→x yy−xxy−x=

Use L-hospital ruleddyyy=yy(1+log⁡y) and ddyxx=0limy→x yy−xxy−x=limy→xyy(1+log⁡y)-01-0=xx(1+logx)

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$\lim_{y \to x} \frac{y^{y} - x^{x}}{y - x} =$

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$x y^{x - 1}$

$y x^{y - 1}$

$x y$

$x^{x} (1 + \log x)$

answer is D.

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Detailed Solution

Use L-hospital rule

$\begin{aligned} (\frac{d}{d y}) (y^{y}) = y^{y} (1 + \log y) \\ and (\frac{d}{d y}) (x^{x}) = 0 \end{aligned}$

$\lim_{y \to x} \frac{y^{y} - x^{x}}{y - x} = \lim_{y \to x} \frac{y^{y} (1 + \log y) - 0}{1 - 0} = x^{x} (1 + \log x)$

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limy→x yy−xxy−x=

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$\lim_{y \to x} \frac{y^{y} - x^{x}}{y - x} =$