[tex]\lim\limits_{x\rightarrow e}\ln(x\ln x)^{\dfrac{1}{x-e}} = \lim\limits_{x\rightarrow e} \Big(1+\ln(x\ln x)-1\Big)^{\dfrac{1}{x-e}} = \\ \\\lim\limits_{x\rightarrow e}\Big(1+\ln(x\ln x)-\ln e\Big)^{\dfrac{1}{x-e}} = \lim\limits_{x\rightarrow e}\left(1+\ln\Big(\dfrac{x\ln x}{e}\Big)\right)^{\dfrac{1}{x-e}} = \\ \\ [/tex]
[tex]= \lim\limits_{x\rightarrow e}\left(1+\dfrac{1}{\dfrac{1}{\ln\Big(\dfrac{x\ln x}{e}\Big)}}\right)^{\dfrac{1}{x-e}} = \\ \\ = \lim\limits_{x\rightarrow e}\left(1+\dfrac{1}{\dfrac{1}{\ln\Big(\dfrac{x\ln x}{e}\Big)}}\right)^{\dfrac{1}{\ln \Big(\dfrac{x\ln x}{e}\Big)}\cdot \ln\Big(\dfrac{x\ln x}{e}\Big)\cdot \dfrac{1}{x-e}} = \\ \\ \\ [/tex]
[tex]\overset{\text{folosim }\lim\limits_{x\rightarrow x_0}(1+\frac{1}{u})^u = e,\quad u\rightarrow \infty}{=} \quad e^{\lim\limits_{x\rightarrow e}\dfrac{\ln\Big(\dfrac{x\ln x}{e}\Big)}{x-e} = $ $ \\ \\ = e^{\lim\limits_{x\rightarrow e}\dfrac{\Big(\ln(x\ln x)-1\Big)'}{\Big(x-e\Big)'}} = e^{\lim\limits_{x\rightarrow e}\dfrac{\dfrac{\ln x+x\cdot \dfrac{1}{x}}{x\ln x}}{1}} = e^{\lim\limits_{x\rightarrow e}\dfrac{\ln x+1}{x\ln x}}} = \\ \\ = e^{\dfrac{\ln e+1}{e\ln e}}} = e^{\dfrac{2}{e}}[/tex]
