[tex]f(x) = \dfrac{1}{1-e^{\frac{1}{x}}},\quad x\in (0,+\infty) \\ \\ y = mx+n\\\\\lim\limits_{x\to \infty}\Big(\dfrac{1}{1-e^{\frac{1}{x}}}-mx-n\Big) = 0 \\ \\ e^{\frac{1}{x}} = t \Rightarrow \frac{1}{x} = \ln t \Rightarrow x= \dfrac{1}{\ln t} \Rightarrow t\to 1\\ \\ \lim\limits_{t\to 1} \Big(\dfrac{1}{1-t}-\dfrac{m}{\ln t}-n\Big) = 0[/tex]
[tex]\lim\limits_{t\to 1} \Bigg[\dfrac{\ln t - m(1-t)}{(1-t)\ln t}\Bigg]-n = 0\\ \\ \text{Aplicam L'Hopital.} \\ \\ \lim\limits_{t\to 1}\Bigg[\dfrac{\frac{1}{t}+m}{-\ln t +\frac{1-t}{t}}\Bigg]-n = 0 \\ \\ \\\Rightarrow m = -1 \text{ deoarece trebuie sa avem nedeterminare, altfel,}\\ \text{limita ar da infinit (1+m)/0 adica infinit.}[/tex]
[tex]\lim\limits_{t\to 1}\Big[ \dfrac{1-t}{-t\ln t+1-t}\Bigg]-n = 0 \\ \\ \text{Aplicam din nou L'Hopital.}\\ \\ \lim\limits_{t\to 1}\Big[\dfrac{-1}{-\ln t -1-1}\Bigg]-n = 0 \\ \\ \Rightarrow \dfrac{1}{2} - n = 0 \Rightarrow n = \dfrac{1}{2}\\ \\ \\ \Rightarrow \boxed{y = -x+\dfrac{1}{2}}[/tex]
=> B) corect.
