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 ?
 A: Note that as $x\to \infty$, the asymptotic expansion of $e^{1/x}$ is 
$$e^{1/x}=1+\frac1x+\frac1{2x^2}+O\left(\frac1{x^3}\right)$$
Hence, 
$$\begin{align}
\lim_{x\to \infty}\left(\frac{1}{1-e^{1/x}}+x\right)&=\lim_{x\to \infty}\left(\frac{1}{1-\left(1+\frac1x+\frac1{2x^2}+O\left(\frac1{x^3}\right)\right)}+x\right)\\\\
&=\lim_{x\to \infty}\left(x-\frac{x}{1+\frac{1}{2x}+O\left(\frac1{x^2}\right)}\right)\\\\
&=\lim_{x\to \infty}\left(\frac12+O\left(\frac1x\right)\right)\\\\
&=\frac12
\end{align}$$
A: Note that
$$
\lim_{x \to \infty} e^{\frac 1x}= e^0=1,
$$
and that $(0, \infty) \ni x \mapsto 1-e^{\frac 1x}$ is non-positive. 
Therefore 
$$
\lim_{x \to \infty} f(x) = - \infty.
$$
Also, $\lim_{x \to \infty} x = \infty$. Therefore by taking the limit $\lim_{x \to \infty} f(x) +x$ you have an expression of the type $-\infty+\infty$ which can - depending on the situation - take any value in $[-\infty,\infty]$.
For example for $c \in \Bbb R$ define $f(x)=-x+ c$ then you also have that $\lim_{x \to \infty} f(x) = - \infty$ and $\lim_{x \to \infty} f(x)+x = c$.
A: 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  $
$= \frac{1}{2} + \mathcal{O}(\frac{1}{x}) \xrightarrow[x \to \infty]{} \frac{1}{2}$
A: In cases like these you can easily get tricked up, so it can be better to do $x=1/t$, whereupon the first limit becomes
$$
\lim_{x\to\infty}\frac{1}{1-e^{1/x}}=
\lim_{t\to0^+}\frac{1}{1-e^t}=-\infty
$$
because $1-e^t<0$ for $t>0$.
Also
$$
\lim_{x\to\infty}(f(x)+x)=
\lim_{t\to0^+}\left(\frac{1}{1-e^t}+\frac{1}{t}\right)=
\lim_{t\to0^+}\frac{t+1-e^t}{t(1-e^t)}=
\lim_{t\to0^+}\frac{t+1-1-t-\frac{t^2}{2}+o(t^2)}{t(1-1-t+o(t))}=
\frac{1}{2}
$$
