How to show monotonocity and the limit? 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.
 A: Let us show that $r(x):=\frac{x-1}{N^{-1}(\Phi(x))}$ is increasing in $x\ge1$. The condition $r'(x)>0$ can be rewritten as 
$y n(y)>k(x-1)n(x-1)$, where $y:=y(x):=N^{-1}(\Phi(x))$ and $k:=\frac{\sqrt{2\pi}}{2+\sqrt{2\pi}}\in(0,1)$. You already know that $N(x-1)<\Phi(x)$, which is equivalent to $y>x-1$. So, it remains to show that $n(y)>k n(x-1)$, which can be rewritten as $y<z(x):=\sqrt{(x-1)^2-2\ln k}$ and then as $\Phi(x)<N(z(x))$ or as $1-k+k N(x-1)<N(z(x))$. Differentiating both sides of the latter inequality, we see that it is enough to show that $k n(x-1)>n(z(x))z'(x)$, which follows because $k n(x-1)=n(z(x))$ and $z'(x)=(x-1)/z(x)<1$. 
To show that $r(x)=\frac{x-1}y\to1$, recall the well-known relation $1-N(x)\sim n(x)/x$, which implies $1-N(x)=\exp\{-\frac12\,x^2\,(1+o(1))\}$;  everything in this paragraph is taken as $x\to\infty$. The definition $y:=N^{-1}(\Phi(x))$ can be rewritten as $N(y)=\Phi(x)$ and then as $1-N(y)=k(1-\Phi(x))$, which shows that $y\to\infty$ and $\exp\{-\frac12\,y^2\,(1+o(1))\}=\exp\{-\frac12\,x^2\,(1+o(1))\}$. So, $y\sim x$; that is, $r(x)=\frac{x-1}y\to1$. 
