Let me reformulate my recent question.

Let $n, N$ denote density and cdf of Gaussian distribution. Let us consider its modification, given by density:

$$\phi(x) = C\left\{ \begin{array}{lcc} \sqrt{e} & \mbox{ for } & x \in [0,1]\\ \exp\left( \frac{-(x-1)^2+1}{2} \right) & \mbox{ for } & x>1\\ \end{array} \right. $$

$$ C= \frac{1}{\sqrt{2\pi e}+2\sqrt{e}}$$ And cdf: $$\Phi(x) = \frac{1}{2} + \int_0^x\phi(s)ds = \left\{ \begin{array}{lcc} C(\frac{\sqrt{2 \pi e}}{2} +\sqrt{e}(x+1))& \mbox{ for } & x \in [0,1]\\ 1-C\sqrt{2\pi e}N(1-x)& \mbox{ for } & x>1\\ \end{array} \right. $$

Im interested in showing that $\frac{\phi(x)}{n(N^{-1}(\Phi(x)))}$ is increasing.

For $x\in[0,1]$ it is obvious.

For $x>1$ we may rewrite $\Phi$ as

$$\Phi(x) = \frac{2}{\sqrt{2\pi}+2}+\frac{\sqrt{2\pi}}{\sqrt{2\pi}+2}N(x-1).$$

Therefore $\frac{N(x−1)}{\Phi(x)}\nearrow1$.

**My question is how to show that $\frac{x−1}{N^{-1}(\Phi(x))}\nearrow1$?**

This fact, coupled with L'Hospital's Monotone Rule, would prove the monotonocity of my function.