Answer
$+\infty$
Work Step by Step
$$\lim_{x \to 0^{+}}xe^{\frac{1}{x}}$$
Rewrite the function:
$$\lim_{x \to 0^{+}}\frac{e^{\frac{1}{x}}}{\frac{1}{x}}=\frac{+\infty}{+\infty}$$
Since it is an indeterminate form type $\frac{\infty}{\infty}$ so using the l'Hospital's rule it follows:
$$\lim_{x \to 0^{+}}\frac{e^{\frac{1}{x}}}{\frac{1}{x}}=\lim_{x \to 0^{+}}\frac{(\frac{1}{x})'e^{\frac{1}{x}}}{(\frac{1}{x})'}=\lim_{x \to 0^{+}}e^{\frac{1}{x}}=+\infty$$
