Evaluating some limits $$f\left(x\right)\:=\:\frac{1}{1-e^{\frac{1}{x}}}, x \in (0, \infty )$$
As far as i know, $\lim _{x\to \infty }f\left(x\right)$ is $\frac{1}{+0}=\infty$. But, $\lim _{x\to \infty }\left(f\left(x\right)\:+\:x\right)=\frac{1}{2}$. How so ?

As mentioned, $\frac{1}{1-e^{\frac{1}{x}}} \xrightarrow[x \to \infty]{}  -\infty$. As for the limit of $f(x) + x$, 
$\frac{1}{1-e^{\frac{1}{x}}} + x = \frac{1}{1-(1+\frac{1}{x} +\frac{1}{2x^2} +\mathcal{O}(\frac{1}{x^3}))} + x$
$= \frac{-x}{1+\frac{1}{2x} + \mathcal{O}(\frac{1}{x^2})} + x = -x(1-\frac{1}{2x} + \mathcal{O}(\frac{1}{x^2})) + x  $
$= \frac{1}{2} + \mathcal{O}(\frac{1}{x}) \xrightarrow[x \to \infty]{} \frac{1}{2}$
