Let $\tilde{x} \sim F(\cdot)$ be a random variable supported on $\text{supp}(\tilde{x}) = [\underline{x}, \overline{x}]$ (for clarity: $\overline{x} < \infty$). Furthermore, let the CDF $F$ be continuosly differentiable everywhere on $[\underline{x}, \overline{x}]$. Define $$ h(x) := \frac{f(x)}{1-F(x)},$$ where $f = F'$ is the pdf of $\tilde{x}$. Is it true that $\lim_{x \to \overline{x}} h(x) = \infty$? I am looking for a proof or a counterexample.
Attempt: I tried to prove this using the fact that $$h(x) = -\frac{\partial}{\partial x}\log(1-F(x),$$ but L'Hopital's rule (applied to $\frac{\log(1-F(x))}{x}$) does not apply because the form is not indeterminate as $x \to \overline{x}$. Furthermore, I attempted to find a counterexample by letting $f'$ go to zero very quickly, but e.g. $f(x) = \frac{11}{10}(1-x^{10})$ with associated CDF $F(x) = \frac{11 x}{10}-\frac{x^{11}}{10}$ still yields $\lim_{x \to \overline{x}} h(x) = \infty$.
It is true that $h(x)$ is unbounded as $x\to\overline x$; for if $h(x)\le M$ then $$-\log(1-F(x)) = \int_{\underline x}^x h(u)\,du\le M(x-\underline x)$$ and the LHS would be finite as $x\to\overline x$.
However, you need additional conditions, for example, that $h$ is increasing, to conclude that $h(x)\to\infty$ as $x\to\overline x$.
A pathological counterexample: On $[0,1]$ let the density $f$ be a sawtooth function consisting of triangles all of height $2$, with corresponding bases on $[0,\frac12], [\frac12,\frac34], [\frac34,\frac78] , \ldots$. So the $n$th triangle has base $[1-\frac1{2^{n-1}},1-\frac1{2^n}]$. You can check that for the sequence $x_n:=1-\frac1{2^n}$ we have $f(x_n)=0$ and $1-F(x_n)=\frac1{2^n}$, hence $x_n\to1$ but $h(x_n)\to0$ as $n\to\infty$. Nonetheless, $h$ is unbounded as $x\to1$.