I have the following to prove :
Let $\phi$ and $\Phi$ denote the standard normal density and distribution functions respesctively. Prove that $$\lim_{x\to\infty}{1-\Phi(x)\over \phi(x)/x}=1$$
I am not being able to start. $\Phi$ is basically $\int \phi(x)dx$. How to calculate the limit?
METHODOLOGY $1$: Application of L'Hospital's Rule
We have $\phi(x)=\frac{1}{\sqrt {2\pi}}e^{-x^2/2}$ and $\Phi(x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty}^x e^{-t^2/2}\,dt$.
Note that $0\le \Phi(x)\le 1$ and $\phi'(x)=-x\phi(x)$.
Therefore, applying L'Hospital's Rule, we have
$$\begin{align} \lim_{x\to \infty}\frac{1-\Phi(x)}{\phi(x)/x}&=\lim_{x\to \infty}\frac{-\phi(x)}{-\phi(x)\left(1+\frac1{x^2}\right)}\\\\ &=-\lim_{x\to \infty}\frac{1}{1+\frac1{x^2}}\\\\ &=1 \end{align}$$
Therefore, we assert that
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\frac{1-\Phi(x)}{\phi(x)/x}=1}$$
as was to be shown!
METHODOLOGY $2$: Using Integration by Parts and the Squeeze Theorem
I thought it might be instructive to present a way forward that forgoes appeal to L'Hospital's Rule. To that end, we proceed.
First, we use integration by parts with $u=\frac1t$ and $v=-\phi(t)$ to obtain
$$\begin{align} 1-\Phi(x)&=\int_x^\infty \phi(t)\,dt\\\\ &=\int_x^\infty \frac1t (t\phi(t))\,dt\\\\ &=\frac{\phi(x)}{x}-\int_x^\infty \frac{\phi(t)}{t^2}\,dt\tag 1 \end{align}$$
From $(1)$ we have the bounds
$$\frac{\phi(x)}{x}-\frac{\phi(x)}{3x^3}\le 1-\Phi(x)\le \frac{\phi(x)}{x} \tag 2$$
Using $(2)$ reveals
$$1-\frac{1}{3x^2}\le \frac{1-\Phi(x)}{\phi(x)/x}\le 1 \tag 2$$
whereupon applying the squeeze theorem to $(3)$ yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\frac{1-\Phi(x)}{\phi(x)/x}=1}$$
