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:
$$
 \lim_{x \to \infty}\left( \frac{x-1}{e^{1/x}} - x \right) = -2.
$$
 A: Set $1/x=u\implies u\to0$
$$\lim_{x \to \infty}\left( \frac{x-1}{e^{1/x}} - x \right) =\lim_{u\to0}\left(\dfrac{1-u}{ue^u}-\dfrac1u\right)$$
$$=\left(\lim_{u\to0}\dfrac1{e^u}\right)\cdot\left(-1-\lim_{u\to0}\dfrac{e^u-1}u\right)$$
Can you take it from here?
A: 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*}
A: Let $\frac 1x=t\implies t\to 0$ as $x\to \infty$
$$\lim_{x\to \infty}\left(\frac{x-1}{e^{1/x}}-x\right)$$
$$=\lim_{t\to 0}\left(\frac{\frac{1}{t}-1}{e^{t}}-\frac{1}{t}\right)$$
$$=\lim_{t\to 0}\left(\frac{1-t-e^t}{te^{t}}\right)$$
$$=\lim_{t\to 0}\left(\frac{\underbrace{1-t-e^{t}}_{\longrightarrow 0}}{\underbrace{te^{t}}_{\longrightarrow 0}}\right)$$
Applying L'Hospital's rule for $\frac 00$ form 
$$=\lim_{t\to 0}\left(\frac{\frac{d}{dt}(1-t-e^{t})}{\frac{d}{dt}(te^{t})}\right)$$
$$=\lim_{t\to 0}\left(\frac{-1-e^{t}}{te^{t}+e^{t}}\right)$$
$$=\frac{-1-e^0}{0+e^0}=\color{red}{-2}$$
A: Take $n=\frac{1}{x}$ and as $x\to\infty$, $n\to 0$.
Therefore, $$\lim_{x\to\infty} \left(\frac{x-1}{e^{\frac{1}{x}}}-x\right)$$
$$\lim_{n\to 0} \left(\frac{\frac{1}{n}-1}{e^n}-\frac{1}{n}\right)$$
$$\lim_{n\to 0} \left(\frac{1-n-e^n}{ne^n}\right) \ldots [\text{of the form} \frac{0}{0}] $$
$$\lim_{n\to 0} \left(\frac{-1-e^n}{e^n+ne^n}\right)$$
$$\left(\frac{-1-e^0}{e^0+0}\right)$$
$$=-2$$
