[tex]f(x) = \dfrac{1-\ln x}{x^2},\quad f(e) = 0\\ \\ \lim\limits_{x\to e} \dfrac{f(x)}{x-e} = \lim\limits_{x\to e} \dfrac{f(x)-f(e)}{x-e} = f'(e) \\ \\ f'(x) = \dfrac{(1-\ln x)'x^2-(1-\ln x)(x^2)'}{x^4} = \dfrac{-\dfrac{1}{x}\cdot x^2-2x(1-\ln x)}{x^4} =\\ \\ = \dfrac{-x-2x(1-\ln x)}{x^4} \\ \\ f'(e) = \dfrac{-e}{e^4} = -\dfrac{1}{e^3}\\ \\\Rightarrow\boxed{\lim\limits_{x\to e} \dfrac{f(x)}{x-e} = -\dfrac{1}{e^3}}[/tex]
