(1) let g(x)=∫[0, x] f(t)dt, then f (x) = x/t * g (t)-g (x) = [x * g (t)-t * g (x)]/.

Because f(x) takes t as the period and f(x+T)=f(x),

So g(x+T)=∫[0, x+T]f(t)dt=∫[0, x] f(t)dt+∫[x, x+T]f(t)dt,

For the latter integral, y=t-x is substituted as a variable, where ∫[x, x+T]f(t)dt=∫[0, T]f(y+x)dy=g(T).

Therefore, g(x+T)=g(x)+g(T),

So f (x+t) = [(x+t) * g (t)-t * (g (x)+g (t))]/t = [x * g (t)-t * g (x)]/t = f (x),

Therefore, F(x) is a periodic function with period t.

(2) g(x)=x/T*g(T)-F(x) is given by (1),

So lim (x→ +∞) g (x)/x = g (t)/t-lim (x→ +∞) f (x)/x = g (t)/t =1/t * ∫ [0, t] f (t).

Where lim(x→+∞) F(x)/x =0 is bounded, because F(x) is a continuous periodic function on R.