For periodic $f,g$ (and continuous), does $\lim (f-g) = 0 $ imply $f=g$? 
Take $f, g: \mathbb{R} \to \mathbb{R} $ continuous and periodic and assume $\lim_{x \to
 \infty} (f(x)-g(x)) = 0$. Does it follow $f=g$? Prove or give counter
  example.

I cant find any counterexample, so maybe we should prove it:
we given that for any $\epsilon > 0$ we can find some $\delta > 0$ such that $x> \delta$ imply $|f(x)-g(x)| < \epsilon $
Let $H(x) = f(x) - g(x)$. and so $H$ better also be periodic and continous.
Now, certainly for any $\delta > 0$ we have $|H|< \epsilon $ which implies $H=0$ or that $f(x) =g(x)$. 
But, we still need to prove for $x \leq \delta$
Now, this seems intuitively obvious as because periodicity and continuity:
if after some $\delta > 0$, we have that $f(x) = g(x)$ then we can find some interval after $\delta> 0$, say $[a,b]$ for which $H(x)$ is identical in $[0,\delta]$ and thus $H(x) = 0$ for every $x \in \mathbb{R}$. Now, this argument seems fishy. 
Am I on the right track or I am completely off tracks?
 A: If I correctly understand your reasoning, you're not far off. However I would put the proof the following way.
If $f$ and $g$ are periodic with the same period $T$ (as you seem to assume in your reasoning), then $H$ is also periodic with period $T$. For any $x\in\mathbb{R}$ we have $\lim_{n\rightarrow\infty} (x+nT) = \infty$. Since $\lim_{x\rightarrow\infty} H(x) = 0$, that means that
$$\lim_{n\rightarrow\infty} H(x+nT) = 0$$
However, because $H$  is periodic, it means that $H(x+nT)=H(x)$ and 
$$\lim_{n\rightarrow\infty} H(x) = 0$$
that is
$$H(x) = 0$$
.
If periods of $T_1$ and $T_2$ of functions $f$ and $g$ are different, but $T_1/T_2 \in\mathbb{Q}$, then you can find $T$ such that $T=nT_1=mT_2$ with $n,m\in\mathbb{N}$. $H$ will be periodic with period $T$, and the proof continues as in the previous case.
If $T_1/T_2\notin\mathbb{Q}$, then there exist two sequences sequences $n_k,m_k\rightarrow\infty$, such that $$\lim_{k\rightarrow\infty} \left(n_kT_1/T_2 - m_k\right) = 0 $$
that is$$\lim_{k\rightarrow\infty} \left(n_kT_1 - m_k T_2\right) = 0 $$
Since $g$ is continuous, we have, for every $x\in\mathbb{R}$:
$$ \lim_{k\rightarrow\infty} g\left(x+n_kT_1 - m_kT_2 \right)= g(x)$$
Then we have
$$0 =\lim_{k\rightarrow\infty} H(x+n_kT_1) = \\
= \lim_{k\rightarrow\infty} \left(f(x+n_kT_1) - g(x+n_kT_1)\right) = \\
= \lim_{k\rightarrow\infty} \left(f(x) - g(x+n_kT_1-m_kT_2) \right) = \\
= f(x)-g(x)$$
.
A: The claim is expressed in a simpler way as $h(x)$ is periodic and has the limit $0$ only if it is identically zero.
This is immediate because if it is not identically zero, it will forever alternate between two values apart.

Addendum:
If the periods are different, we have two cases:


*

*the periods are commensurable: in this case $h$ is still periodic;

*the periods are incommensurable: in this case $h$ is aperiodic but $n_1T_1-n_2T_2$ will come arbitrarily close to $0$ and using the continuity property, the alternation between values apart remains.
A: Suppose $f, g$ have non-zero periods $T_1, T_2$ respectively and $h=f-g, R = T_2/T_1$ We have 2 cases:
$R = p/q \in \mathbb{Q}:$ Then $h$ is periodic with period $T = pT_1,$ so $h(x) = \lim\limits_{n \to \infty} h(x) = \lim\limits_{n \to \infty} h(x+nT) = \lim\limits_{x \to \infty} h(x) = 0.$
$R \not\in \mathbb{Q}.$ Suppose $x \in \mathbb{R}.$ We can pick a sequence $\{x_n\} \in \mathbb{N}$ such that $T_2 x_n \mod T_1 \to 0$ as $n \to \infty.$ Then $0 = \lim\limits_{x \to \infty} h(x) = \lim\limits_{n \to \infty} h(x+T_2 x_n) = h(x)$ by the continuity of $h.$ 
So in either case, $h \equiv 0 \Rightarrow f = g.$
A: Let the periods of $f$ and $g$ be $T_1$ and $T_2$ respectively. Let $H(x) = f(x) - g(x)$. 
$H(x+ T_1) - H(x) = f(x+T_1) -g(x+T_1) - f(x) + g(x) = g(x) - g(x+T_1)$. So we have 
$$ \lim_{x\to\infty}H(x+T_1) - H(x) = 0 - 0 = 0 =  \lim_{x\to\infty}\left(g(x) - g(x+T_1)\right)$$
Note that $g(x) - g(x+T_1)$ is a periodic function, now one can adopt the proof @Adam Latosiński suggested, to show that $g(x) = g(x+T_1)$, i.e. $f$ and $g$ should have the same period. So $H$ also has the period $T_1$, and @Adam Latosiński's proof can be applied again for $H$, since now we know that $H$ is periodic (of period $T_1$).
