$-\log\Phi(x) \sim \frac{x^2}{2}$ as $x \to -\infty$ for $\Phi(x) = (2\pi)^{-\frac 12}\int_{-\infty}^x \mathrm{e}^{-t^2/2}\,\mathrm{d}t$ For $$\Phi(x) = (2\pi)^{-\frac 12}\int_{-\infty}^x \mathrm{e}^{-t^2/2}\,\mathrm{d}t,$$ it is claimed in the proof of Lemma 8.12 of this book that we have the asymptotic relation $-\log\Phi(x) \sim \frac{x^2}{2}$ as $x \to -\infty$. However, I do not see any way of deriving it from the so-called Gordon's inequality, which says that $$1 - \Phi(t) \sim \frac{1}{t\sqrt{2\pi}}\mathrm{e}^{-t^2/2}$$ for $t \gg 1$ being sufficiently large. May I know how can we obtain the advertised asymptotic relation as $x \to -\infty$?

Let $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. Notice that
$$\begin{align}
-\frac{d}{dx}\Big(\Big(\frac1x-\frac1{x^3}\Big)\phi(x)\Big)&=(1-3x^{-4})\phi(x)\\
&\leq  \phi(x)\leq (1+x^{-2})\phi(x)=-\frac{d}{dx}\Big(\frac{\phi(x)}{x}\Big)
\end{align}$$
For $a<0$, after tntegrating over $(-\infty,a]$, we obtain
$$\Big(\frac1{a^2}-\frac1a\Big)\phi(a)\leq \Phi(a)\leq-\frac{1}{a}\phi(a)$$
and so
$$\lim_{a\rightarrow-\infty}\frac{-a\Phi(a)}{\phi(a)}=1$$
which (using L'Hospital rule for example) gives
$$\lim_{a\rightarrow-\infty}\frac{-\log(\Phi(a))}{a^2/2}=1$$
