Infinity - infinity calculus

Question

$$\lim_{x\to\infty} (x-1)e^{-1/x}-x$$

I know that this limit equals $-2$ but I don't know how to prove it. I can only get to $\infty-\infty=?$

Answer 1

$$(x-1)e^{-1/x} - x = x(e^{-1/x} - 1) - e^{-1/x}$$

$$\lim x(e^{-1/x} - 1) = \lim \frac{e^{-1/x} - 1}{1/x}$$

Use L'hopital's rule.

Answer 2

You can use $e^{-1/x} = 1 - \tfrac 1x + O(\tfrac 1{x^2})$:

$$\begin{align} \lim_{x\to\infty} (x-1)e^{-1/x}-x & = \lim_{x\to\infty} (x-1)(1 - \tfrac 1x + O(\tfrac 1{x^2}))-x \\ &= \lim_{x\to\infty} (x-1) + -\tfrac 1x(x-1) + O(\tfrac 1{x^2})-x \\ &= \lim_{x\to\infty} x-1 + -1 +\tfrac 1x + O(\tfrac 1{x^2})-x \\ &= \lim_{x\to\infty} -2 +\tfrac 1x + O(\tfrac 1{x^2})\\ &= -2 \end{align}$$