Sufficient Condition for Asymptotically Exponential Tail? Corrected but still unsolved conjecture! I conjecture a sufficient condition for a distribution to be asymptotically exponential in a strong sense.
Roughly speaking, the idea is this. Suppose the "expected residual lifetime," $E[X-x|X≥x]$ is approximately constant for large $x$. Then, I believe that the conditional tail distribution is approximately exponential, in the sense of being stochastically dominated by an exponential and dominating a similar exponential. Formally:
Conjecture Given any random variable  $X$ with support on $[0,∞)$. If $$lim_{x→∞}E[X-x|X≥x]= \lambda ,$$
then for all $ε>0$ and for all $\Delta>0$ there is some $c$ such that $x≥c$ implies  $$e^{-(1/(λ-ε))t}≥Pr[X≥x+t|X≥x]≥e^{-(1/(λ+ε))t}    \qquad ∀t≥\Delta.$$
Update 2 The first conjecture was wrong. Robert Israel provided a counterexample. The implication is now weaker, restricting $∀t≥\Delta>0$. The weakening takes care of the counterexample. But is it correct?
Update 3 (Removed) 
Update 4 I posted the question on MathOverflow.
Update 5 The approximation result is stronger than weak convergence. Let $Y$ be distributed exponentially with parameter $\lambda$. The conclusion of the conjecture implies that
$$\lim_{x→∞}\mathbb{E}[f(X-x)|X≥x]=\mathbb{E}[f(Y)]$$
for all nondecreasing functions for which $\mathbb{E}[f(Y)]$ exists. In particular, $f$ is allowed to be unbounded.

What happens as $t \to 0+$?  $1 - e^{-t/(\lambda \pm \epsilon))} \approx \frac{t}{\lambda \pm \epsilon}$, so your conclusion would imply that if $x'$ is sufficiently large, the density $f(x')$ exists and
$$\frac{1}{\lambda + \epsilon} \le \frac{f(x')}{1-F(x')} \le \frac{1}{\lambda - \epsilon}$$
That is too strong a conclusion, I think.  For example, you could try a discrete distribution with possible values, say, $x_n = \sqrt{n}$ for positive integers $n$, and $P(X = x_n) = e^{-x_{n-1}} - e^{-x_n}$ with $P(X=x_1) = 1- e^{-x_1}$.  The CDF of this distribution agrees with that of the exponential distribution of parameter $1$ at the points $x_n$, and I think it's easy to see that it will satisfy your hypothesis.
