If $\Phi$ is the standard normal cumulative distribution function, what is the value of this expression
$$\lim_{x\rightarrow \infty} {x (1-\Phi(x))}$$ as $x$ approaches infinity?
While $x$ tends to infinity, the term $(1-\Phi(x))$ tends to zero , so it is not clear if it has a limit.
Use de l'Hôpital to transform it into
$$\lim_{x\to\infty}x(1-\Phi(x))=\lim_{x\to\infty}\frac{1-\Phi(x)}{1/x}\overset{H}{=}\lim_{x\to\infty}\frac{-\phi(x)}{-1/x^2}=\lim_{x\to\infty}x^2\phi(x)=\frac{1}{\sqrt{2\pi}}\lim_{x\to\infty}x^2\exp(-x^2/2) \; .$$
Can you work this out from this point forward?