Function being equal on infinitely many points implies it's constant? I have the following problem and solution given in my textbook:
Does there exist a nonconstant function $f : (1,∞) → \mathbb{R}$ satisfying the relation $f(x)= f(\frac{x^2+1}{2} )$ for all $x>1$ and such that $lim_{x\rightarrow \infty}f(x)$exists? 
For $x>1$ deﬁne the sequence $(x_n)n≥0$ by $x_0 = x$ and $x_{n+1} = x^2_{n}+1/2$ , $n ≥ 0$. The sequence is increasing because of theAM–GM inequality. Hence it has a limit L, ﬁnite or inﬁnite. Passing to the limit in the recurrence relation, we obtain $L^2 -2L +1 = 0$ ; hence either $L =1$ or $L =∞$. Since the sequence is increasing, $L ≥ x0 > 1$, so $L =∞$. We therefore have $f(x)= f(x_0) = f(x_1) = f(x_2) =···= lim_{n→∞} f(x_n) = lim_{x→∞} f(x)$. This implies that $f$ is constant, which is ruled out by the hypothesis. So the answer to the question is negative.
I am having trouble with the last part, why does the values that the function takes on at a certain set of points imply it is constant for all points because in general, this is not true.
 A: Define $C=lim_{x\rightarrow \infty}f(x)$. You proved that for any $x$ in the domain of $f$, we have $f(x)=C$. And that is the definition of a constant function.
That's it. Job done!
It's possible that you lost sight of your progress because you start the proof with "For $x>1$" but you then reuse the letter $x$ as the variable in the limit. If you avoid reusing letters, you can avoid that sort of confusion.
Edit: Let me be clearer. I am not investigating whether a constant function satisfies the given relations. Obviously it does. I am describing what you have already proved.
Your proof starts with "For $x>1$", without making any further assumptions about $x$. This means that any conclusions you draw apply to all such $x$. You then prove the sequence of equations $f(x)= f(x_0) = f(x_1) = f(x_2) =···= \lim_{n\to\infty} f(x_n) = \lim_{x\to\infty} f(x)$. In particular, the left-hand side is equal to the right-hand side: $f(x)=\lim_{x\to\infty} f(x)$. But the right-hand side is not a function of $x$; it is a constant. We can name this constant $C$. So you have successfully proved that $f(x)=C$ for all $x$. In other words, $f$ is a constant function.
In summary, you have proved that if $f$ satisfies the given relations, then it is constant. So the answer to the question is negative.
