Answer
$\infty$
Work Step by Step
Here, we have $\lim\limits_{x \to 0} f(x)=\lim\limits_{x \to \infty}
xe^{1/x}$
But $\lim\limits_{x \to 0} \dfrac{e^{1/x}}{\frac{1}{x}}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{x \to 0} \dfrac{e^{1/x}(\frac{-1}{x^2})}{\frac{-1}{x^2}}=\lim\limits_{x \to 0} e^{1/x}$
or, $e^{1/0}=\infty$
