Limit of $\frac{1 - F(x)}{f(x)}$ Suppose $f: [0, 1] \rightarrow \mathbb{R}+$ is a continuous density function. Let $F$ be its c.d.f. Suppose $f > 0$ over $(0, 1)$ and $$\frac{1 - F(x)}{f(x)}$$ is decreasing. Is it true that $$\lim_{x \rightarrow 1} \frac{1 - F(x)}{f(x)} = 0?$$
Trivially, if $f(1) > 0$, then it is true. But how about $f(1) = 0$?
 A: The assertion is true. The reciprocal function to $\frac{1-F(x)}{f(x)}$ is the hazard function
$$h(x):=\frac{f(x)}{1-F(x)}.\tag1$$
Check that $h$ satisfies
$$
\int_0^x h(u)\,du=-\log[1-F(x)]\tag2
$$
for every $x$. In particular, as $x$ approaches $1$ from below,
$$
\lim_{x\to1}\int_0^x h(u)\,du=\infty.\tag3
$$
Reason: For $x\in(0,1)$ we know $f(x)>0$, so $F(x)$ increases to $1$ as $x$ approaches $1$, hence the RHS of (2) approaches infinity.
Note that $h$ is increasing (since by assumption its reciprocal is decreasing), so (3) implies that $h(x)\to\infty$ as $x\to1$; otherwise the integral $\int_0^1 h(u)du$ would be finite. As a consequence we see
the reciprocal function must decrease to $0$.
A: if the distribution is over $[0,1]$ for $f$ then it makes sense that the cumulative probability, $F(1)=1$ since this is the upper limit of pdf, meaning that:
$$\lim_{x\to 1}1-F(x)=0$$
However, if we do not know anything about the type of pdf $f$ is or its skewness how can be know the value of $f(x)$ at a given point, and therefore the value of $f(1)$? Without this I dont think it is possible to fully conclude the value of the limit given.
