how to solve this problem $\lim_{\lambda\to0}\lambda \,log(e^{\xi_1/\lambda}+\ldots+e^{\xi_n/\lambda})$ I find this problem
$$\lim_{\lambda\to0}\lambda \,log(e^{\xi_1/\lambda}+\ldots+e^{\xi_n/\lambda})$$ where $\xi_1, \ldots, \xi_n, \lambda \in \mathcal{R}$ how to solve it?
 A: The limit $\lambda \to 0$ doesn't exists unless $\xi_1 = \xi_2 = \cdots = \xi_n$.
To see this, let us consider the two sided limits. 
Let 
$\xi_{max} = \max(\xi_1,\xi_2,\ldots,\xi_n)$ and
$\xi_{min} = \min(\xi_1,\xi_2,\ldots,\xi_n)$.
When $\lambda > 0$, we have 
$$\xi_1 
= \lambda\log(e^{\xi_{max}/\lambda})
\le \lambda\log(e^{\xi_1/\lambda} + e^{\xi_2/\lambda} + \cdots + e^{\xi_n/\lambda})
\le \lambda\log(n e^{\xi_{max}/\lambda})
= \xi_{max} + \lambda\log n
$$
Since $\lim\limits_{\lambda\to 0} \lambda\log n = 0$, this implies the limit
from the right exist:
$$\lim\limits_{\lambda\to 0^+}\lambda\log(e^{\xi_1/\lambda} + e^{\xi_2/\lambda} + \cdots + e^{\xi_n/\lambda}) = \xi_{max}$$
By a similar argument, one find the limit from the left also exist:
$$\lim\limits_{\lambda\to 0^-}\lambda\log(e^{\xi_1/\lambda} + e^{\xi_2/\lambda} + \cdots + e^{\xi_n/\lambda}) = \xi_{min}$$
Unless $\xi_1 = \xi_2 = \cdots = \xi_n$, we have
$$\lim_{\lambda\to 0^+}(\cdots) = \xi_{max} \ne \xi_{min} = 
\lim_{\lambda\to 0^-}(\cdots)
\quad\implies\quad
\lim_{\lambda\to 0}(\cdots) \text{ doesn't exist }.
$$
A: Let us look inside the sum  in two cases: 
(a) $\lambda$ > 0: when $\lambda$ approaches $0$, only the terms with $\xi >0$ 'survive' (the rest go to $0$). Let us agree for this case that we refer here onwards in the sum only to the positive $\xi$ be default. Then:
$$ \lim_{\lambda \to 0} \frac{\ln \sum e^{\xi_i/\lambda}}{1/ \lambda} =\lim_{\lambda \to 0}\frac{\sum \xi_i e^{\xi_i/\lambda} }{\sum e^{\xi_i/\lambda}} =\xi_{max}$$ (you can see the last equality if you split the sum into the one term containing $\xi_{max} \, $ and the rest).
(b) $\lambda<0$: just change the sign of $\lambda$ and we have the previous case (a), only that instead now the result is $\xi_{min}$ ($=-\vert \xi_{min}\vert$).
As the previous solution of @achille-hui says in the end, the limit exists only if the minimum and maximum $\xi$ are identical (implying all $\xi$ are equal).
