$-\log\Phi(x) \sim \frac{x^2}{2}$ 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 $x \to -\infty$?

I try to turn the way around: for $x\to +\infty$, $\log ( (2\pi)^{-\frac 12}\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t )\sim -x^2/2$. Since this is asymptotic, we omit the $(2\pi)^{-\frac 12}$ and only focus on the integral.
Notice that
$$\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t\ge \int_{x}^{x+1/x} \mathrm{e}^{-t^2/2}\,\mathrm{d}t\ge\int_{x}^{x+1/x} \mathrm{e}^{-(x+1/x)^2/2}\,\mathrm{d}t=\frac 1x\mathrm{e}^{-\frac{x^2+2+1/x^2}{2}}.$$
So when taking logarithms, we have $\log(\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t)\ge-\frac{x^2}{2}+O(\log x)$.
Furthermore, notice that for $t\ge x$, we have $t^2\ge x^2+2x(t-x)$. So we have, for $x>0$,
$$\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t\le \int_{x}^{\infty} \mathrm{e}^{\frac{-x^2-2x(t-x)}{2}}\,\mathrm{d}t=\mathrm{e}^{\frac{-x^2}{2}}\int_{x}^{\infty} \mathrm{e}^{-x(t-x)}\,\mathrm{d}t=\mathrm{e}^{\frac{-x^2}{2}}\int_{0}^{\infty} \mathrm{e}^{-xt}\,\mathrm{d}t=\mathrm{e}^{\frac{-x^2}{2}}\frac{1}{x}.$$
So when taking logarithms, we also have $\log(\int_{x}^\infty \mathrm{e}^{-t^2/2}\,\mathrm{d}t)\le-\frac{x^2}{2}+O(\log x)$.
Therefore $\log \Phi(x)\sim -\frac{x^2}{2}$. What's more, we can prove that $\log \Phi(x)= -\frac{x^2}{2}-\log x+O(1)$.
