The difference between two periodic functions converges to zero, is this two functions identical? If $f(x)$ and $g(x)$ are two periodic functions, that is, $f(x+T_1)=f(x)$ and $g(x+T_2)=g(x)$ for every $x \in \Bbb R$. Now that $\displaystyle\lim_{x\to\infty}(f(x)-g(x))=0$.
Conjecture: $f(x) \equiv g(x)$.
 A: Let $h(x)=f(x)-g(x)$. The assumption $h\to 0$ implies $h(x+T_2)-h(x)\to 0$ as $x\to \infty$. Since  $h(x+T_2)-h(x) = f(x+T_2)-f(x)$, it follows that 
$$f(x+T_2)-f(x)\to 0 \quad x\to\infty \tag{1}$$  
As a corollary of (1), for any fixed $x$ we have 
$$\lim_{n\to\infty} (f(x+nT_1+T_2)-f(x+nT_1)) = 0$$  On the other hand, the expression in parentheses is independent of $n$. Thus, the expression must equal $0$ for all $n$, so $f$ is $T_2$-periodic. It follows that $f-g$ is also $T_2$-periodic. A periodic function with limit $0$ at infinity must be identically $0$. 
A: I wanted to add a little more explanation about how the other answer concludes $f(x)$ is $T_2$-periodic, starting from
$$
f(x+T_2)-f(x)\to 0\tag{$*$}
$$
Fix $x$, and let $\epsilon>0$. Because of $(*)$, there exists an $M$ so $|x|>M$ implies $|f(x+T_2)-f(x)|<\epsilon$. Therefore, choosing $n$ large enough so $|x+nT_1|>M$,
$$
|f(x+T_2)-f(x)|=|f(x+nT_1+T_2)-f(x+nT_1)|<\epsilon
$$
Because $|f(x+T_2)-f(x)|<\epsilon$ for all $\epsilon>0$, it follows $f(x+T_2)-f(x)=0$. Therefore, you get $f$ is $T_2$ periodic, and can conclude as in the other proof.
A: $$
f(x) = \lim_{n\to\infty}f(x)-f(x+nT_1)+g(x+nT_1) = \lim_{n\to\infty}g(x+nT_1).$
Also we get $
g(x) = \lim_{n\to\infty}f(x+nT_1).$
Then $f(x)-g(x) = \lim_{n\to\infty}g(x+nT_1+nT_2)-f(x+nT_2+nT_1)=0$.
