Evaluating an infinity/infinity limit using L'Hopital's Rule

First, rewrite $x^2 e^{-x} = \dfrac{x^2}{e^x}$. As $x \to \infty$, both $x^2 \to \infty$ and $e^x \to \infty$, so this is an $\infty/\infty$ indeterminate form, and L'Hopital's Rule applies. Differentiating numerator and denominator separately gives $\lim_{x\to\infty} \frac{x^2}{e^x} = \lim_{x\to\infty} \frac{2x}{e^x}.$

The new limit $\dfrac{2x}{e^x}$ is again of the form $\infty/\infty$, so L'Hopital's Rule can be applied a second time: differentiate numerator and denominator again to get $\lim_{x\to\infty} \frac{2x}{e^x} = \lim_{x\to\infty} \frac{2}{e^x}.$ As $x \to \infty$, $e^x \to \infty$, so this limit equals $0$. Therefore $\lim_{x\to\infty} x^2 e^{-x} = 0$.

L'Hopital's Rule was applied twice in part a. The first application turned x squared over e to the x into two x over e to the x, which was still infinity over infinity. The second application turned that into two over e to the x, which directly evaluated to zero. So two applications were needed before the indeterminate form was resolved.