In footnote 17 of Lauermann and Wolinsky, "Search with adverse selection" Econometrica (Jan 2016) http://www.jstor.org/stable/43866544, they note the following "known result" about the limit of the inverse hazard rate:
For any random variable $Y$ with unbounded support and c.d.f. $G(y)$ and p.d.f. $g(y)$, $$\lim_{y\to\infty}\frac{1-G(y)}{g(y)}=λ \implies \lim_{x\to\infty}\mathbb{E}(Y-x\,|\,Y≥x)=\lambda $$ For the exponential distribution this holds away from the limit as well.
How can this be shown? They also note that $\frac{y}{-\log(1-G(y))}$ has the same limit as the inverse hazard rate (as can be seen using L'Hôpital's rule), but I don't see how this can be used to imply the limiting case of the above conditional expectation.
Note that $$\mathbb{E}\left[Y-x\middle|Y\geq x\right]=\frac{\int_{y\geq x}\left(y-x\right) dG\left(y\right)}{1-G\left(x\right)}.$$ Both denominator and numerator go to zero as $x\to\infty$, apply L'Hôpital's rule and Leibniz rule, it gives $$\lim_{x\to\infty}\mathbb{E}\left[Y-x\middle|Y\geq x\right]=\lim_{x\to\infty}\frac{0-\int_{y\geq x}1dG\left(y\right)}{-g\left(x\right)}=\lambda.$$