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[Гост]

[tex]\lim_{x \to 1}[/tex] [tex](x-1)^{lnx }[/tex]

[Davids]

Нека $L := \lim_{x\to 1}(x-1)^{\ln x}$

Тогава $\ln L = \lim_{x\to 1}\ln(x-1)\ln(x) = \lim_{x\to 1}\frac{\ln(x-1)}{\frac{1}{\ln x}} \stackrel{L'Hopital}{=}\lim_{x\to 1}\frac{\frac{1}{x-1}}{-\frac{1}{x\ln^2x}} = -\lim_{x\to 1}\frac{x\ln^2x}{x-1} \stackrel{L'Hopital}{=} -\lim_{x\to 1}\Bigl(\ln^2x + 2\ln x\Bigr) = 0$

И така $L = 1$.

[pipi langstrump]

$\lim_{x\to 1}\ln(x-1)\ln(x) = \lim_{y\to 0} \ln(y) \ln(1+y)$ $ =\lim_{y\to 0} y\ln y = \lim_{z\to +\infty} - \frac{\ln z}{z} =0$