Necessary likelihood ratios conditions for stochastic dominance Suppose $X$ has CDF and PDF $F_X$ and $f_X$ with support $(-\infty,\infty)$ and $Y$ has CDF and PDF $F_Y$ and $f_Y$ also with support $(-\infty,\infty)$.

Muller (2001) claims that if $X$ is stochastically dominated by $Y$ (i.e. $F_X(t)\geq F_Y(t)$ for all $t$) then
  $$
\lim_{t\to-\infty}\frac{f_Y(t)}{f_X(t)}\leq 1,\quad \lim_{t\to+\infty}\frac{f_Y(t)}{f_X(t)}\geq 1.
$$
  Why is this true please?

I understand the intuition of the result considering $F_X\geq F_Y$ and 
$$
0=\lim_{t\to-\infty}F_X(t)=\lim_{t\to-\infty}F_Y(t),\quad 1=\lim_{t\to+\infty}F_X(t)=\lim_{t\to+\infty}F_Y(t).
$$
But a rigorous argument escapes me.
 A: Suppose $\lim_{t\rightarrow-\infty} \frac{f_Y(t)}{f_X(t)}$ exists. We get: 
$$ \frac{F_Y(t)}{F_X(t)} \leq 1 \: \: \: \: \forall t $$
and so if $\lim_{t\rightarrow-\infty}\frac{F_Y(t)}{F_X(t)}$ exists, it must be less than or equal to 1. But it gives a case of $0/0$, so using L'Hopital's rule we get: 
$$1 \geq \lim_{t\rightarrow-\infty} \frac{F_Y(t)}{F_X(t)} = \lim_{t\rightarrow-\infty} \frac{f_Y(t)}{f_X(t)}$$ 
You can do the same trick with $(1-F_Y(t))/(1-F_X(t))$ to get the limit as $t\rightarrow\infty$ (assuming the limit of the PDFs exists). 

The next question is whether the limit of PDFs even exists, and I do not think it always does. 

To see that it may not exist, you can take $X$ as Gaussian $N(0,1)$.  Then take $F_Y(t)$ as a near-piecewise constant function that is under $F_X(t)$ always, but has an infinite number of constant levels as $t\rightarrow -\infty$.  Then the PDF $f_X(t)$ goes to zero as $t\rightarrow-\infty$, but the PDF $f_Y(t)$ would have consistent spikes near infinity, so $f_Y(t)/f_X(t)$ can be larger than 1000 consistently for arbitrarily small $t$. It cannot have a limit since any limit must be less than or equal to 1. 
