Rewriting and evaluating a product-form limit using L'Hopital's Rule

First rewrite the product as a quotient: $x^2 e^{-x} = \frac{x^2}{e^x}$. As $x \to \infty$, the numerator $x^2 \to \infty$ and the denominator $e^x \to \infty$, so this is the indeterminate form $\infty/\infty$, which justifies applying L'Hopital's Rule.

Differentiating numerator and denominator separately gives $\lim_{x\to\infty} \frac{2x}{e^x}$, which is again $\infty/\infty$, so L'Hopital's Rule is applied a second time: $\lim_{x\to\infty} \frac{2}{e^x}$. As $x \to \infty$, $e^x \to \infty$, so this limit is $0$. Therefore $\lim_{x\to\infty} x^2 e^{-x} = 0$.

The rule was applied twice. First, x squared over e to the x was differentiated to get two x over e to the x, since it was still infinity over infinity. Then that result was differentiated again to get two over e to the x, which gave a determinate limit of zero. Two applications were needed before the indeterminate form disappeared.