Answer
$\infty$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to \infty}
xe^{1/x}$
and
$\lim\limits_{x \to 0} \dfrac{e^{1/x}}{1/x}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{e^{1/x}(-1/x^2)}{-1/x^2}=\lim\limits_{x \to 0} e^{1/x}=\infty$