Mean residual life function limits In my paper about mean residual life functions and heavy tails I've come across the examples listed in the picture below.


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*http://i.stack.imgur.com/OnklD.png

*source (page 124/125): http://www.mi.uni-koeln.de/~schmidli/vorl/Risk/vorl.pdf
I've managed to successfully calculate the limits when $M \rightarrow\infty$ for the Exponential and Pareto distribution myself using the  equation $$\frac{\int_M^\infty {(1-F(z))dz}}{1-F(M)}$$ however I fail to understand how to calculate the remaining three (marked in yellow). I am new to probability so please bear with me. Thank you for your aid.
 A: First, let us write the formulas themselves.  As I come from an actuarial background, my notation is slightly different:  $$\overset\circ e_m = \operatorname{E}[T-m \mid T > m] = \frac{\int_{t=m}^\infty S(t) \, dt}{S(m)} = \int_{t=0}^\infty \vphantom._t p_m \, dt$$ where $T$ is the lifetime random variable, $S(t) = 1 - F(t)$ is the survival function of $T$, and $\vphantom._t p_m = S(m+t)/S(m)$ is the probability of survival of a life aged $m$ to age $m+t$.  The key observation is that as $m \to \infty$, both the numerator and denominator tend to $0$, so L'Hopital's rule applies.  Then by the fundamental theorem, and noting $S(\infty) = 0$:  $$\lim_{m \to \infty} \overset\circ e_m = \lim_{m \to \infty} \frac{-S(m)}{-f(m)} = \lim_{m \to \infty} \frac{-f(m)}{f'(m)},$$ where $f$ is the density of $T$.  Now all that remains is to choose a density:  for a gamma distribution with shape $\gamma$ and rate $\alpha$, we have $$f(m) \propto m^{\gamma-1} e^{-\alpha m},$$ consequently $$f'(m) \propto (\gamma - 1)m^{\gamma-2} e^{-\alpha m} - m^{\gamma - 1} \alpha e^{-\alpha m} = m^{\gamma - 2} e^{-\alpha m} (\gamma - 1 - \alpha m).$$  Hence $$\lim_{m \to \infty} \frac{-f(m)}{f'(m)} = \frac{m}{\alpha m + 1 - \gamma},$$ and as $m \to \infty$, this is clearly $1/\alpha$ as claimed.  A similar approach works for the other two distributions.  If you remain stuck in the calculation, I will amend this answer to include those computations.
