Let $x_1, x_2, \dots, x_n$ be a finite set of real numbers and $w(x, \theta)$ a weight function such that
$$w(x, \theta) = e^{x\theta}.$$
Denote $a(\theta)$ the weighted average of the $x_i$'s with weights given by $w$, i.e., $$a(\theta) = \frac{\sum_{i = 1}^n x_i e^{x_i\theta}}{\sum_{i = 1}^n e^{x_i\theta}}.$$
Intuitively, I think that $\lim_{\theta \rightarrow \infty} = \max_i x_i$ and $\lim_{\theta \rightarrow -\infty} = \min_i x_i$ because when $\theta$ is too large, then the relative weight of the maximum value will make all others negligible. I don't know what tools I can use to prove these limits. Is there any trick I can use here?
Really interesting question hope I can help! This might seem very long and wordy but I have just explained things in a lot of detail, its not that horrible don't worry! Before we begin to answer the question we are going to need to prove a tiny lemma!
$\textbf{Lemma:} $
Let $a_{max}$ = max {$a_1,a_2,a_3,..,a_k$} then $\lim_{n \to +\infty} \frac{a_{max}^n}{a_1^n+a_2^n+a_3^n+...+a_k^n} =1$
We will use the following simple law of limits (*) : $\lim_{n \to +\infty} \frac{1}{s_n} = \frac{1}{\lim_{n \to +\infty} s_n }$
Proof of Lemma:
$\lim_{n \to +\infty}\frac{a_1^n+a_2^n+a_3^n+...+a_k^n}{a_{max}^n} = \lim_{n \to +\infty}((\frac{a_1}{a_{max}})^n + (\frac{a_2}{a_{max}})^n + ... + (\frac{a_{max}}{a_{max}})^n + ...+ (\frac{a_k}{a_{max}})^n )$
(All we have done is split the sum into its $k$ pieces. And there is one $a_i$ such that $a_{i} = a_{max}$, from definition of $a_{max}. $ Yes this $i$ could be $1,2$ or $k$ but this doesn't matter, the expansion is just for a visual aid. )
$\lim_{n \to +\infty}((\frac{a_1}{a_{max}})^n + (\frac{a_2}{a_{max}})^n + ... + (\frac{a_{max}}{a_{max}})^n + ...+ (\frac{a_k}{a_{max}})^n) = \lim_{n \to +\infty}(r_1^n + r_2^n + ... + 1^n + ... + r_k^n)$
Where $r_i := \frac{a_i}{a_{max}}$
Notice that for all i we have $r_i < 1 $
Hence: $\lim_{n \to +\infty} r_i^n = 0 $ $\forall$ $i $
And then $\lim_{n \to +\infty}(r_1^n + r_2^n + ... + 1^n + ... + r_k^n) =$ 0 + 0 + ... + 1 + ... + 0 = $1$
Hence:
$\lim_{n \to +\infty}\frac{a_1^n+a_2^n+a_3^n+...+a_k^n}{a_{max}^n} = 1 $ and then from (*) law we finally have: $\lim_{n \to +\infty} \frac{a_{max}^n}{a_1^n+a_2^n+a_3^n+...+a_k^n} =\frac{1}{1} = 1$
Don't worry! That was the hard part! The rest is easy dont worry! Now onto the last stretch:
Firstly let us define $a_i := e^{x_i} $
$ \lim_{\theta \to +\infty} a(\theta) = \lim_{\theta \to +\infty} \frac{\sum_{i = 1}^n x_i a_i^{\theta}}{\sum_{i = 1}^n a_i^{\theta}} = \lim_{\theta \to +\infty} (\frac{x_{1}a_1^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + \frac{x_{2}a_2^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + ... + \frac{x_{max}a_{max}^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + ... + \frac{x_{n}a_n^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}})$
This is the same trick as before, writing a sum as all of its components but separately listing the maximum. We must now try to evaluate this limit for each component. We will split these into 2 cases to do seperately. Those with $ a_j \not = a_{max} $ and then $a_{max} $
$\textbf{Case 1} $
To evaluate $\lim_{\theta \to +\infty} (\frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}) $ for $ a_j \not = a_{max} :$
Notice that $ \sum_{i = 1}^n a_i^{\theta} $ (the denominator of our components) is greater than $ a_{max}^{\theta} $.
Hence $ \frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} < \frac{x_{j}a_j^{\theta}}{a_{max}^{\theta}} $ as it has a greater denominator
Hence: $\lim_{\theta \to +\infty} \frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} < \lim_{\theta \to +\infty} \frac{x_{j}a_j^{\theta}}{a_{max}^{\theta}} $
Now, like before, define $r_j := \frac{a_j}{a_{max}} $ < 1
And hence: $\lim_{\theta \to +\infty} \frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} < \lim_{\theta \to +\infty} \frac{x_{j}a_j^{\theta}}{a_{max}^{\theta}} = \lim_{\theta \to +\infty} = a_j \cdot r_j^{\theta} = 0 $
So $\lim_{\theta \to +\infty} (\frac{x_{j}a_j^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}) $ for $ a_j \not = a_{max} = 0 $
$\textbf{Case 2} $
$\lim_{\theta \to +\infty} (\frac{x_{max}a_{max}^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}) = \lim_{\theta \to +\infty} x_{max} \cdot \frac{a_{max}^{\theta}}{\sum_{i = 1}^n a_i^{\theta}} = x_{max} \cdot 1 $ (From the Lemma we proved)
Hence:
$ \lim_{\theta \to +\infty} a(\theta) = \lim_{\theta \to +\infty} \frac{\sum_{i = 1}^n x_i a_i^{\theta}}{\sum_{i = 1}^n a_i^{\theta}} = \lim_{\theta \to +\infty} (\frac{x_{1}a_1^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + \frac{x_{2}a_2^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + ... + \frac{x_{max}a_{max}^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}} + ... + \frac{x_{n}a_n^{\theta}}{ \sum_{i = 1}^n a_i^{\theta}}) = 0 + 0 + ... + x_{max} + ... + 0 = x_{max} $
Finally for $ \lim_{\theta \to -\infty} a(\theta) $:
Notice $ \lim_{\theta \to -\infty} a(\theta) = \lim_{\theta \to +\infty} a(-\theta) = \lim_{\theta \to +\infty} \frac{\sum_{i = 1}^n x_i e^{-x_i\theta}}{\sum_{i = 1}^n e^{-x_i\theta}} = -\lim_{\theta \to +\infty} \frac{\sum_{i = 1}^n -x_i e^{-x_i\theta}}{\sum_{i = 1}^n e^{-x_i\theta}}$ =
$-$Max{$-x_1,-x_2,...,-x_n$} = $- - $Min{$x_1,x_2,...,x_n$} = Min{$x_1,x_2,...,x_n$}
Your intuition was correct!
Hope this helped.
Oskar