Prove that f(x)=g(x) 
Show that if $f,g:\mathbb{R}\to \mathbb{R}$ are continuous and periodic and $\lim_{x\to \infty}[f(x)-g(x)]=0$,
  then $f=g$

 A: Let $T_1, T_2 > 0$ be periods of $f$ respectively $g$. 
We break the problem in  two cases:
Case 1: $\frac{T_1}{T_2}$ is rational. Let $m,n$ be integers so that 
$$\frac{T_1}{T_2}=\frac{m}{n} \,.$$
Then $f(x)-g(x)$ is periodic with period $T=mT_2=nT_1$. The conclusion follows easily from here:
$$f(x)-g(x)=f(x+kT)-g(xkT) $$
and let $k \to \infty$.
Case 2: $\frac{T_1}{T_2}$ is irrational.
Then the set $\{ mT_1+nT_2 \}$ is dense in $\mathbb R$. 
Then you can find a sequence $\alpha_k=m_kT_1-n_kT_2$ so that $m_k, n_k \to \infty$ and  $\alpha_k \to 0$.
Then, for all $x \in \mathbb R$ we have
$$f(x)-g(x+\alpha_k)=f(x+m_kT_1)-g(x+\alpha_k+n_kT_2)= \left( f-g \right)(x++m_kT_1)\,.$$
Now, by letting $k \to \infty$ and using the continuity of $g$ you get 
$$f(x)-g(x)=0 \,.$$
A: Let $a,b$ be (positive) periods of $f$ and $g$, respectively.
If $\frac ab$ is rational, say $\frac ab=\frac nm$ with $n,m\in\mathbb N$, then $p:=ma=nb$ is a common period of $f$ and $g$. Let $x_0\in\mathbb R$. Then $x_0+kp\to\infty$ implies $|f(x_0)-g(x_0)|=|f(x_0+kp)-g(x_0+kp)|\to 0$, hence $f(x_0)=g(x_0)$ as was to be shown.
If $\frac ab$ is irrational, then we can find good rational approximations, that is infinitely many pairs $(n,m)$ with $m$ arbitrarily large and $\left|\frac ab-\frac nm\right|<\frac 1{m^2}$.
Let $x_0\in\mathbb R$.
Then for $\epsilon>0$ we find $\delta>0$ such that $|f(x)-f(x_0)|<\frac12\epsilon$ whenever $|x-x_0|<\delta$. Also, by assumption $|f(x)-g(x)|<\frac \epsilon2$ for $x$ sufficiently large, i.e. for approximations with sufficiently large $m$ we have
$$ \begin{align}|g(x_0)-f(x_0)|&\le|g(x_0+nb)-f(x_0+nb)|+|f(x_0+nb-ma)-f(x_0)|\\&<\frac12\epsilon+\frac12\epsilon=\epsilon\end{align}$$
Since $\epsilon>0$ was arbitrary, $f(x_0)=g(x_0)$.
A: Assume for simplicity that the period is $1$.  The formula $\lim_{x\to \infty}f(x)-g(x)=0$ means that for an infinite value of $x$, say $x=H$, the difference $f(H)-g(H)$ is infinitesimal, or $f(H)-g(H)\approx 0$.  By periodicity, the same is true for the fractional part $h=\{H\}$, namely $f(h)-g(h)\approx 0$.  Now let $h_0=\text{st}(h)$ be its standard part.  By continuity, we still have $f(h_0)-g(h_0)\approx 0$.  But the values $f(h_0)$ and $g(h_0)$ are real.  Since the only real infinitesimal is $0$, we obtain $f(h_0)-g(h_0)=0$. The answer follows from the fact that every number in $[0,1)$ is the fractional part of a suitable infinite number.
Now assume that $f$ has period $1$ while $g$ has a different period $\alpha$. Since fractional parts of the integer multiples of $\alpha$ are dense in $[0,1]$, we can choose a hyperinteger $H$ such that $\{H\alpha\}\approx 0$, in other words $H\alpha$ is infinitely close to a hyperinteger $K$.
Then for any real $r$, we have $f(r)=f(r+K)\approx f(r+H\alpha)\approx g(r+H\alpha)=g(r)$.  Thus $f(r)\approx g(r)$, and since both are real, they must be equal as above.
A: Lets first prove the following claim

Claim : If $h : \mathbb{R} \to \mathbb{R}$ is continuous, periodic and $\lim_{x \to \infty}h(x) = 0$ then $h$ is zero function

Proof : Let the period of $h$ be $a \;$  i.e.  $h(x+a)=h(x) \; \; \forall x$
$\lim_{x \to \infty}h(x) = 0 \implies \lvert h(x) \rvert < \epsilon \text{   for } x>x_0$
By translating back the function, we see that $\lvert h(x) \rvert < \epsilon \; \; \forall x$ . As $\epsilon $ is arbitrary, we can claim that $h$ is identically 0.
Now the required claim follows by using that if $f$ and $g$ are continuous and periodic then $f-g$ is continuous and periodic.
