Evaluate the limit: lim (x->-infinity) x2 * e^x

As $x \to -\infty$, $e^x$ approaches 0, and $x^2$ approaches $\infty$.
The limit is of the indeterminate form $0 \times \infty$.

To use L'Hôpital's Rule, we need the expression in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. We can rewrite $e^x \times x^2$ as $\frac{x^2}{e^{-x}}$.

As $x \to -\infty$, the numerator $x^2 \to \infty$ and the denominator $e^{-x} \to e^{-(-\infty)} = e^{\infty} = \infty$. This is an $\frac{\infty}{\infty}$ form.

As $x \to -\infty$, the numerator $2x \to -\infty$ and the denominator $-e^{-x} \to -e^{-(-\infty)} = -\infty$. This is still an $\frac{\infty}{\infty}$ form.

As $x \to -\infty$, $e^{-x}$ approaches $e^{-(-\infty)} = e^{\infty} = \infty$.
Therefore, the expression $\frac{2}{e^{-x}}$ approaches $\frac{2}{\infty}$, which is 0.