The difference between two periodic functions converges to zero, is this two functions identical?

Question

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: $f(x) \equiv g(x)$.

Answer 1

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 $0$.

Answer 2

$f(x) = \lim_{n\to\infty}f(x)-f(x+nT_1)+g(x+nT_1) = \lim_{n\to\infty}g(x+nT_1).$

Also we get $ g(x) = \lim_{n\to\infty}f(x+nT_2).$

Then $f(x)-g(x) = \lim_{n\to\infty}g(x+nT_1+nT_2)-f(x+nT_2+nT_1)=0$.