The following simple equation takes in an N-length (real) vector, and spits out a (real) number between 0 and 1. (I believe this means that it is a transformation mapping $\mathfrak{R}^N \rightarrow \mathfrak{R}^1$). This equation has the property that the answer will converge to 1, as the input elements become more and more alike.
Anyway, here is the equation. $\bf x$ is the N-dimensional input vector. $y$ is the 1-dimensional scalar output. $e$ is just the exponent operator.
$$ y = \frac{e^{\frac{1}{N}\displaystyle\sum_{n=1}^N \log_e(x[n])}}{\frac{1}{N}\displaystyle\sum_{n=0}^N x[n]} $$
(Notice how the denominator is just the mean of $\bf x$). This function will return $y=1$ if all the elements of the vector $x$ are equal to each other.
I am trying to do two things:
1) First, I would like to 'translate' the English statement "Show that this equation converges to 1, as all the elements in the vector $\bf x$ become more and more alike", into a mathematical statement. (Italics stressing what I want translated). For example, in English we might say "As the variable $c$ approaches infinity, etc, and we use the $\displaystyle\lim_{c \to +\infty}$ to mathematically denote that.
2) After that, I would like to show (prove), that the above equation actually does converge to 1, as the elements of the above equation $x$ become more and more similar. (I know that it does indeed converge to 1, but would like to prove it).
First of all, note that in order to get real values, you want all $x_n > 0$. The equation can be simplified as $$ y = \dfrac{\prod_{n=1}^N x_n^{1/N}}{\frac{1}{N} \sum_{n=1}^N x_n}$$ You could say that for any $c > 0$, $$ \lim_{{\bf x} \to [c,\ldots,c]} \dfrac{\prod_{n=1}^N x_n^{1/N}}{\frac{1}{N} \sum_{n=1}^N x_n} = 1 $$ This is true because the expression is a continuous function on $(0,\infty)^N$ and, as you noted, the value is $1$ when all $x_n$ are equal to the same positive number.