## Calculus (3rd Edition)

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Since $$\lim _{x \rightarrow \infty} \frac{ x^2 }{e^x}=\frac{\infty}{\infty}$$ is an intermediate form, then we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow \infty} \frac{ x^2 }{e^x}=\lim _{x \rightarrow \infty} \frac{ 2x }{e^x}=\frac{\infty}{\infty}.$$ Again, by applying L’Hôpital’s Rule, we get $$\lim _{x \rightarrow \infty} \frac{ 2x }{e^x}=\lim _{x \rightarrow \infty} \frac{ 2 }{e^x}=\frac{2}{\infty}=0.$$