Limit of e^x/x^2 as x approaches infinity using L'Hopital's Rule

As $x \to \infty$, both $e^x \to \infty$ and $x^2 \to \infty$, so the limit has the indeterminate form $\infty/\infty$. This justifies applying L'Hopital's Rule: differentiate the numerator and denominator separately (not using the quotient rule) to get $\lim_{x\to\infty} \frac{e^x}{x^2} = \lim_{x\to\infty} \frac{e^x}{2x}.$

The new limit $\lim_{x\to\infty} \frac{e^x}{2x}$ is again of the form $\infty/\infty$, so L'Hopital's Rule applies once more: differentiating top and bottom gives $\lim_{x\to\infty} \frac{e^x}{2x} = \lim_{x\to\infty} \frac{e^x}{2}.$ As $x \to \infty$, $e^x \to \infty$, so this limit is $\infty$, meaning the original limit diverges to infinity.