Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 52

Determine the $\displaystyle \lim_{x \to \infty} \left( xe^{\frac{1}{x}} - x \right)$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to \infty} \left( xe^{\frac{1}{x}} - x \right) = \lim_{x \to \infty} x\left( e^{\frac{1}{x}} - 1 \right) = \lim_{x \to \infty} \frac{\left(e^{\frac{1}{x}} - 1 \right)}{\frac{1}{x}}$

By applying L'Hospital's Rule...

$\begin{equation} \begin{aligned} \lim_{x \to \infty} \frac{\left(e^{\frac{1}{x}} - 1 \right)}{\frac{1}{x}} &= \lim_{x \to \infty} \frac{e^{\frac{1}{x}}\left( \frac{-1}{x^2} \right) }{\left( \frac{-1}{x^2} \right)}\\\ \\ &= \lim_{x \to \infty} e^{\frac{1}{x}}\\ \\ &= e^{\frac{1}{\infty}}\\ \\ &= e^0\\ \\ &= 1 \end{aligned} \end{equation}$