= Solution
Write $Q(x)=1-\Phi(x)=\int_x^\infty\phi(t)\,dt$, where $\phi(t)=(2\pi)^{-1/2}e^{-t^2/2}$ is the <standard normal density>. For $x>0$, $\phi'(t)=-t\phi(t)$ and <integration by parts> yield
$$
Q(x)=\int_x^\infty\frac{-\phi'(t)}t\,dt
=\frac{\phi(x)}x-\int_x^\infty\frac{\phi(t)}{t^2}\,dt.
$$
The boundary term at infinity vanishes. The remainder is nonnegative and is at most $Q(x)/x^2$, so
$$
\frac{x\phi(x)}{x^2+1}\leq Q(x)\leq\frac{\phi(x)}x.
$$
Dividing by the positive quantity $\phi(x)/x$ gives lower and upper bounds $x^2/(x^2+1)$ and $1$. The <squeeze theorem> proves the <Mills ratio> asymptotic
$$
\boxed{1-\Phi(x)\sim\frac{\phi(x)}x\qquad(x\to\infty).}
$$
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