limit of logarithm function containing finite sum of exponential functions Let $f(x)=\frac{1}{x} \log{[\sum_{i=1}^{n}e^{xa_{i}}]}$. Prove that $\lim\limits_{x\rightarrow -\infty}f(x)=\min_{1\leq i \leq n}a_{i}$!
Okay, so I have tried to solve this problem using squeeze theorem:

For $x<0$, we will get $\frac{1}{x} \log{[\sum_{i=1}^{n}e^{x\min_{1\leq i \leq n}a_{i}}]} = \frac{1}{x} \log{[ne^{x\min_{1\leq i \leq n}a_{i}}]} = \frac{1}{x}[\log(n)+ x\min_{1\leq i \leq n}a_{i}]\geq f(x)$

Therefore, we can get this inequality:
$\lim\limits_{x\rightarrow -\infty}f(x) \leq \lim\limits_{x\rightarrow -\infty}\frac{1}{x}[\log(n)+ x\min_{1\leq i \leq n}a_{i}]$
$\lim\limits_{x\rightarrow -\infty}f(x) \leq \min_{1\leq i \leq n}a_{i}$


The problem is, I can`t find the left part of the inequality to use the squeeze theorem.
Can anyone give me a hint / clue of what to do with the left side so that I can get $\min_{1\leq i \leq n}a_{i}\leq\lim\limits_{x\rightarrow -\infty}f(x) \leq \min_{1\leq i \leq n}a_{i}$?
Thank you very much! Any help is much appreciated!
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 So, after trying to figure the solution, I came up with this idea:

Observe that for $x<0$ we have:
$e^{xa_{i}} > 0$ and $e^{x \min_{1\leq i \leq n}a_{i}} \geq e^{xa_{i}}$
Therefore, we can conclude that
$e^{x \min_{1\leq i \leq n}a_{i}} \leq \sum_{i=1}^n[e^{xa_{i}}] \leq \sum_{i=1}^n[e^{x \min_{1\leq i \leq n}a_{i}}]$
$e^{x \min_{1\leq i \leq n}a_{i}} \leq \sum_{i=1}^n[e^{xa_{i}}] \leq n e^{x \min_{1\leq i \leq n}a_{i}}$

By taking the logarithm value of the inequality since logarithmic function is strictly increasing, we get the following inequality:
$\log e^{x \min_{1\leq i \leq n}a_{i}} \leq \log(\sum_{i=1}^n[e^{xa_{i}}]) \leq \log(n e^{x \min_{1\leq i \leq n}a_{i}})$
Multiplying the inequality with $\frac{1}{x}<0$, we get a new inequality:
$\frac{1}{x}\log(n e^{x \min_{1\leq i \leq n}a_{i}}) \leq \frac{1}{x}\log(\sum_{i=1}^n[e^{xa_{i}}]) \leq \frac{1}{x}\log(n e^{x \min_{1\leq i \leq n}a_{i}})$
$\frac{1}{x}\log(n) + \min_{1\leq i \leq n}a_{i} \leq \frac{1}{x}\log(\sum_{i=1}^n[e^{xa_{i}}]) \leq \min_{1\leq i \leq n}a_{i}$
Taking the limit of the inequality above as $x \to -\infty$ will result in:
$\lim\limits_{x\rightarrow -\infty}\frac{1}{x}\log(n) + \min_{1\leq i \leq n}a_{i} \leq \lim\limits_{x\rightarrow -\infty}\frac{1}{x}\log(\sum_{i=1}^n[e^{xa_{i}}]) \leq \lim\limits_{x\rightarrow -\infty}\min_{1\leq i \leq n}a_{i}$
$\min_{1\leq i \leq n}a_{i} \leq \lim\limits_{x\rightarrow -\infty}\frac{1}{x}\log(\sum_{i=1}^n[e^{xa_{i}}]) \leq \min_{1\leq i \leq n}a_{i}$
Finally, applying the squeeze theorem, we have:
$\lim\limits_{x\rightarrow -\infty}\frac{1}{x}\log(\sum_{i=1}^n[e^{xa_{i}}]) = \min_{1\leq i \leq n}a_{i}$
 So, this is my attempt of the solution. Did I make a mistake in some of the steps?
 A: Assume $0<a_1<a_2<\cdots <a_n.$ We have that
$$\lim_{x\to -\infty} \dfrac{\sum_{i=1}^n a_ie^{xa_i}}{\sum_{i=1}^n e^{xa_i}}=\lim_{x\to -\infty} \dfrac{\sum_{i=1}^n a_ie^{x(a_i-a_1)}}{\sum_{i=1}^n e^{x(a_i-a_1)}}=a_1,$$ since $$\lim_{x\to -\infty}e^{x(a_i-a_1)}=0, \quad i=2,\cdots, n.$$
So, if we apply L'Hôpital's rule to the original limit we get the answer.
A: I may contribute partially, hope this helps. 
We may use L'Hospital here, and see that we need to find the limit of  : 
 $$ \lim_{x \rightarrow -\infty} g(x) = \lim_{x \rightarrow -\infty} \frac{\sum a_{i}e^{a_{i}x}}{\sum e^{a_{i}x}} $$
Now let
$$  y = e^{x} $$
then we can write
$$ \lim_{x \rightarrow -\infty} \frac{\sum a_{i}e^{a_{i}x}}{\sum e^{a_{i}x}}  = \lim_{y \rightarrow 0} \frac{\sum a_{i}y^{a_{i}}}{\sum y^{a_{i}}}   $$
Let $a_{M}$ be minimum of $a_{i}$'s, if there are $k$ of these (there might be more than one mininum) then
$$ \lim_{y \rightarrow 0} \frac{\sum a_{i}y^{a_{i}}}{\sum y^{a_{i}}}  = \lim_{y \rightarrow 0} \frac{k a_{M} + \sum_{a_{i} \ne a_{M}} a_{i}y^{a_{i}-a_{M}}}{k + \sum_{a_{i} \ne a_{M}} y^{a_{i}-a_{M}}}  $$ 
You may conclude from here.
