Limit of normal hazard rate I'm trying to work out the asymptotic behavior of the normal hazard rate as $x$ gets very large.  To be clear, that's the behavior of 
$$ h(x) = \frac{ \phi(x)}{1-\Phi(x)} \qquad \text{ as } \qquad x \rightarrow \infty$$
Where $\phi(\cdot)$ and $\Phi(\cdot)$ are the pdf and cdf respectively of the standard normal.
I don't think anyone on this site or elsewhere online has addressed this question specifically.  The closest I found was this, but that is much more general than what I want.
In the limit, this looks like it is linear (picture below), but I can't quite show why or to what limit.  Both numerator and denominator go to 0 as $x$ gets very large.
My best attempt at figuring this out was to apply L'Hopital's rule.
\begin{align*}
  \lim_{x\rightarrow\infty} h(x) &= \frac{ \lim_{x\rightarrow\infty} \phi'(x) }{ \lim_{x\rightarrow\infty} -\phi(x)} \\
                                 &= \frac{ \lim_{x\rightarrow\infty} -x \phi(x) }{ \lim_{x\rightarrow\infty} -\phi(x)} 
\end{align*}
This still fails as both the top and bottom converge to zero.    The limits don't exist, so I'm no good here.
Now I know that the next line is not ok, but I tried it anyway, because I had no better ideas.  What if I "cancel" the $\phi(\cdot)$ functions in the numerator and denominator?  That is, I tried the following:
$$ \lim_{x\rightarrow\infty} h(x) \overset{?}{=} x$$
It turns out that this is a pretty good approximation to the limiting behavior. See the picture below ($h(x)$ solid, candidate $h'(x)$ dashed).  But I haven't proved anything, which is annoying.  It also turns out that this limit doesn't work in some other applications (not discussed here!).

So, to summarize my questions:


*

*What is the asymptotic behavior of the normal hazard rate?  I couldn't find a reference.

*Is $h'(x)=x$ in the limit?

*If so, why?  My abuse of L'Hopital's rule isn't the reason.

 A: A lower bound for the hazard is $h(s)>s,s>0$. I have established an upper bound, which might be new.
For $s>0$,
$$
h\left( s\right) <\frac{1}{%
s}+s.
$$
Proof: 
Note that
$$
-\left[ \phi _{0}\left( s\right) \frac{s}{1+s^{2}}%
\right] ^{\prime }=\phi _{0}\left( s\right) \left[ 1-2\frac{1}{%
\left( s^{2}+1\right) ^{2}}\right]
$$
and that
$$
1-2\frac{1}{\left( t^{2}+1\right) ^{2}}<1.
$$
Then
$$
\phi _{0}\left( s\right) \frac{s}{1+s^{2}}%
=\int_{s}^{\infty }\phi _{0}\left( t\right) \left( 1-2\frac{1}{%
\left( t^{2}+1\right) ^{2}}\right) dt<1-\Phi_{0}\left(
s\right) .
$$
The conclusion follows since $s>0$ implies that $\frac{s}{1+s^{2}}>0$.
Remark This improves the bound $h(s)<\frac{s^3}{s^2 -1}$ provided by Feller (1957) Lemma 2, Chapter VII.
A: It suffices to consider the standard normal distribution, as any other univariate normal with mean $\mu$ and standard deviation $\sigma$ can be standardized by a suitable location-scale transformation.  Thus we have $$f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}, \quad -\infty < x < \infty,$$ and $$F(x) = \int_{z=-\infty}^x f(z) \, dz.$$  Then the hazard rate is $$h(x) = \frac{f(x)}{1-F(x)},$$ and the limit by L'Hopital's rule is $$\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \frac{f'(x)}{-f(x)},$$ as you observed.  Now $$f'(x) = -x f(x),$$ hence we simply have $$\lim_{x \to \infty} h(x) = \lim_{x \to \infty} x = \infty.$$  It is easy to see that the hazard is asymptotic to $x$ for large $x$.  It is not too difficult to obtain an asymptotic series expansion about $x = \infty$:  $$h(x) = x + \frac{1}{x} - \frac{2}{x^3} + \frac{10}{x^5} + O(x^{-7}).$$  Perhaps surprisingly, $$h(x) \approx \frac{x + \sqrt{x^2+4}}{2}$$ is also an excellent approximation, certainly better for "small" $x$ than the asymptotic series above.
