$\lim_{x\to\infty}{1-\Phi(x)\over \phi(x)/x}=1$ 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?
 A: 
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

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\frac{1-\Phi(x)}{\phi(x)/x}=1}$$

