[tex]\displaystyle \\ \lim\limits_{x \to \infty} \frac{1}{1-e^{\dfrac{1}{x}}} +x =? \\ \\ $Facem schimbare de variabila: t = \dfrac{1}{x} \Rightarrow t \rightarrow \dfrac{1}{\infty} \Rightarrow t\rightarrow 0 \\ x = \dfrac{1}{t}; \\ \Rightarrow \lim\limits_{t\rightarrow0} \Big(\dfrac{1}{1-e^t}+ \dfrac{1}{t}\Big) =\lim\limits_{t\rightarrow0} \dfrac{t+1-e^t}{t(1-e^t)} \overset{\frac{0}{0}(L'H.)}{=} \lim\limits_{t\rightarrow0} \dfrac{1-e^t}{1-e^t+t\cdot(-e^t)} = [/tex]
[tex]\overset{\frac{0}{0}(L'H.)}{=} \lim\limits_{t\rightarrow0} \dfrac{-e^t}{-e^t-e^t+t(-e^t)} = \dfrac{-e^0}{-e^0-e^0+0(-e^0)} = \dfrac{-1}{-1-1+0} = \\ \\ =\dfrac{-1}{-2} = \dfrac{1}{2} \\ \\ \Rightarrow \boxed{\boxed{\lim\limits_{x \to \infty} \frac{1}{1-e^{\dfrac{1}{x}}} +x =\dfrac{1}{2}}}[/tex]
