What does this notation mean? $f(x) \equiv A$? If $\lim_{x\rightarrow\infty}f(x) = A$, prove that $f(x)\equiv A$.
Any help would be appreciated! I just don't even understand the question.
 A: For any $a\in (0,\infty)$, the constant $f(a)=f(2^n a)\to A$ as $n\to\infty$. Done. 
A: Further thoughts and a little problem:
Interesting. This is similar to (but not the same as) a periodic function. 
In a periodic function f(x + k) = f(x) for all x. For example in f(x) = sin(x), the minimum k = 2$\pi$ so the period is 2$\pi$.
Here we have f(cx) = f(x) with c = 2. The "periods" increase geometrically, with f(x) = f(2x) = f(4x) = f(8x) $\ldots$
One example of such a function would be $$f(x) = sin(\frac{2\pi}{log(2)} log ( x))$$
Then $$f(2x) = sin(\frac{2\pi}{log(2)} log( 2x))$$
$$ = sin(\frac{2\pi}{log(2)} (log( 2) +log(x)))$$ 
$$ = sin( 2\pi +\frac{2\pi}{log(2)}log(x))$$
$$ = sin( \frac{2\pi}{log(2)}log(x))$$
$$ = f(x)$$
This is more detail than strictly required by the question but it is an interesting pattern and worth exploring.
In the above function the value of f(x) keeps varying between 1 and -1 even as the "periods" get longer exponentially. If as given in the question $\lim_{x\rightarrow \infty} f(x) = A$ then the above function would not work.
But now what would be wrong with this function, which approaches zero as x tends to infinity? 
$$ f(x) = e^{-x} sin( \frac{2\pi}{log(2)}log(x))$$
The above function is defined on $(0, \infty)$ as required [If we were required to use a closed interval that would be a problem for log but as given above the interval is open] and it seems to satisfy the requirements given. But it is not constant.
Thoughts anyone?
