Two indeterminate limits via L'Hopital

Evaluate (a) $\displaystyle\lim_{x\to\infty}\frac{e^x}{x^2}$ and (b) $\displaystyle\lim_{x\to0^+}x^2\ln x$.

(a) $+\infty$; (b) $0$. (a) Apply L'Hopital twice: $e^x/x^2\to e^x/(2x)\to e^x/2\to\infty$. (b) Rewrite as $\frac{\ln x}{x^{-2}}$; one application gives $\frac{1/x}{-2x^{-3}}=-\tfrac{x^2}{2}\to0$.