limy→xyy−xxy−x=

Using L-Hospital rule, (ddy)(yy)=yy(1+logy)and(ddy)(xx)=0limy→x yy(1+logy)1=xx(1+logx)

limy→xyy−xxy−x=

Using L-Hospital rule, (ddy)(yy)=yy(1+logy)and(ddy)(xx)=0limy→x yy(1+logy)1=xx(1+logx)

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

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$x y^{x - 1}$

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

$y . x^{y - 1}$

answer is B.

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Detailed Solution

Using L-Hospital rule, $(\frac{d}{d y}) (y^{y}) = y^{y} (1 + \log y) a n d (\frac{d}{d y}) (x^{x}) = 0$

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

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