The stated result can be easily obtained by successive applications of integration by parts. We know that $1 - \Phi(x) = \Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} e^{-t^2/2}dt$. To apply integration by parts, multiply and divide the integrand by  $\frac{1}{t}$ to obtain $\Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \frac{1}{t} \left(t e^{-t^2/2}\right)dt$.

The terms shown in brackets can be integrated to obtain $-e^{-t^2/2}$. Hence, the integral simplifies to $\Phi^c(x) = \frac{1}{\sqrt{2\pi}} \frac{e^{-x^2/2}}{x} - \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2}$.

We know that $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$. Also, the integral term on the right hand side is always greater than 0. More specifically, $\frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2} \geq 0$. Hence, combining these observations we find that $\Phi^c(x) \leq \frac{\phi(x)}{x}$ or in other words, $\frac{1 - \Phi(x)}{\phi(x)} \leq \frac{1}{x}$.

Now, you could continue to apply integration by parts again on $\Phi^c(x)$ to obtain a lower bound. More specifically, we obtain $\Phi^c(x) = \phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) + \int_{x}^{\infty} 3t^{-4} e^{-t^2/2}dt$. Again, the integral on the right hand side is positive and hence, $\phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) \leq \Phi^c(x)$.

Combining both these inequalities we find that $\frac{1}{x} - o(x^{-1}) \leq \frac{\Phi^c(x)}{\phi(x)} \leq \frac{1}{x}$. In other words, $\frac{\Phi^c(x)}{\phi(x)} \approx \frac{1}{x}$ in the asymptotic limit of $x$.

It turns out that more applications of integration by parts leads to an alternating series which is enveloping. More details can be referenced from the textbook by Christopher G Small on Expansions and Asymptotics for Statistics which should be available online.

The stated result can be easily obtained by successive applications of integration by parts. We know that $$ 1 - \Phi(x) = \Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} e^{-t^2/2}dt. $$ To apply integration by parts, multiply and divide the integrand by  $\frac{1}{t}$ to obtain $$ \Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \frac{1}{t} \left(t e^{-t^2/2}\right)dt $$ The terms shown in brackets can be integrated to obtain $-e^{-t^2/2}$. Hence, the integral simplifies to $$ \Phi^c(x) = \frac{1}{\sqrt{2\pi}} \frac{e^{-x^2/2}}{x} - \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2}. $$ Now we know that

• $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ and moreover
• the integral term on the right hand side is always greater than 0, i.e. specifically, $$ \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2} \geq 0. $$

Hence, combining these observations we find that $$ \Phi^c(x) \leq \frac{\phi(x)}{x} $$ or in other words, $$ \frac{1 - \Phi(x)}{\phi(x)} \leq \frac{1}{x} $$

Now, you could continue to apply integration by parts again on $\Phi^c(x)$ to obtain a lower bound. More specifically, we obtain $$ \Phi^c(x) = \phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) + \int_{x}^{\infty} 3t^{-4} e^{-t^2/2}dt. $$ Again, the integral on the right hand side is positive and hence, $$ \phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) \leq \Phi^c(x). $$ Combining both these inequalities we find that $\frac{1}{x} - o(x^{-1}) \leq \frac{\Phi^c(x)}{\phi(x)} \leq \frac{1}{x}$. In other words, $$ \frac{\Phi^c(x)}{\phi(x)} \approx \frac{1}{x} $$ in the asymptotic limit of $x$.

It turns out that more applications of integration by parts leads to an alternating series which is enveloping. More details can be taken from the textbook by Christopher G Small, Expansions and asymptotics for statistics, Monographs on Statistics and Applied Probability 115. Boca Raton, FL: CRC Press, ISBN 978-1-58488-590-0/hbk; 978-1-4200-1102-9/ebook, pp. xiv+343 (2010), MR2681183, Zbl 1196.62002, which should be available online.

The stated result can be easily obtained by successive applications of integration by parts. We know that $1 - \Phi(x) = \Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} e^{-t^2/2}dt$. To apply integration by parts, multiply and divide the integrand by $\frac{1}{t}$ to obtain $\Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \frac{1}{t} \left(t e^{-t^2/2}\right)dt$.

The terms shown in brackets can be integrated to obtain $-e^{-t^2/2}$. Hence, the integral simplifies to $\Phi^c(x) = \frac{1}{\sqrt{2\pi}} \frac{e^{-x^2/2}}{x} - \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2}$.

We know that $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$. Also, the integral term on the right hand side is always greater than 0. More specifically, $\frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2} \geq 0$. Hence, combining these observations we find that $\Phi^c(x) \leq \frac{\phi(x)}{x}$ or in other words, $\frac{1 - \Phi(x)}{\phi(x)} \leq \frac{1}{x}$.

Now, you could continue to apply integration by parts again on $\Phi^c(x)$ to obtain a lower bound. More specifically, we obtain $\Phi^c(x) = \phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) + \int_{x}^{\infty} 3t^{-4} e^{-t^2/2}dt$. Again, the integral on the right hand side is positive and hence, $\phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) \leq \Phi^c(x)$.

Combining both these inequalities we find that $\frac{1}{x} - o(x^{-1}) \leq \frac{\Phi^c(x)}{\phi(x)} \leq \frac{1}{x}$. In other words, $\frac{\Phi^c(x)}{\phi(x)} \approx \frac{1}{x}$ in the asymptotic limit of $x$.

The stated result can be easily obtained by successive applications of integration by parts. We know that $1 - \Phi(x) = \Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} e^{-t^2/2}dt$. To apply integration by parts, multiply and divide the integrand by $\frac{1}{t}$ to obtain $\Phi^c(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \frac{1}{t} \left(t e^{-t^2/2}\right)dt$.

The terms shown in brackets can be integrated to obtain $-e^{-t^2/2}$. Hence, the integral simplifies to $\Phi^c(x) = \frac{1}{\sqrt{2\pi}} \frac{e^{-x^2/2}}{x} - \frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2}$.

We know that $\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$. Also, the integral term on the right hand side is always greater than 0. More specifically, $\frac{1}{\sqrt{2\pi}} \int_{x}^{\infty} \frac{e^{-t^2/2}}{t^2} \geq 0$. Hence, combining these observations we find that $\Phi^c(x) \leq \frac{\phi(x)}{x}$ or in other words, $\frac{1 - \Phi(x)}{\phi(x)} \leq \frac{1}{x}$.

Now, you could continue to apply integration by parts again on $\Phi^c(x)$ to obtain a lower bound. More specifically, we obtain $\Phi^c(x) = \phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) + \int_{x}^{\infty} 3t^{-4} e^{-t^2/2}dt$. Again, the integral on the right hand side is positive and hence, $\phi(x) \left( \frac{1}{x} - \frac{1}{x^3} \right) \leq \Phi^c(x)$.

Combining both these inequalities we find that $\frac{1}{x} - o(x^{-1}) \leq \frac{\Phi^c(x)}{\phi(x)} \leq \frac{1}{x}$. In other words, $\frac{\Phi^c(x)}{\phi(x)} \approx \frac{1}{x}$ in the asymptotic limit of $x$.

It turns out that more applications of integration by parts leads to an alternating series which is enveloping. More details can be referenced from the textbook by Christopher G Small on Expansions and Asymptotics for Statistics which should be available online.