Why is $\lim_{x \to \infty}\left( \frac{x-1}{e^{1/x}} - x \right) = -2$? Please, I need the method of doing the following limit:
$$
 \lim_{x \to \infty}\left( \frac{x-1}{e^{1/x}} - x \right) = -2.
$$

Let $y = 1/x$. then
\begin{align*} \lim_{y \to 0} \frac{1/y-1}{e^y} - \frac{1}{y} = \lim_{y \to 0} \frac{1-y-e^y}{ye^y} = \lim_{y \to 0} \frac{-1-e^y}{ye^y + e^y} = -2\end{align*}
