\ln  A=\lim_{x\to 0-} \, x \ln \left(e^{-\frac{1}{x^2}}\right)=\lim_{x\to 0-} \, x \frac{-1}{x^2}=\frac{-1}{x}=+\infty  \Rightarrow A=e^{+\infty }=+\infty