Limit of a piecewise continuous and periodic function

Question

How do I take the following limit:

$$ \lim _{x\rightarrow \infty} F(x)/x $$

where $$ F' = f$$

and $f$ is a piecewise continuous function and 2 pi periodic. I tried $ f = (cosx)^2$ and got 1 for the above limit but how do I show this with a formal argument?

Answer 1

From OP's choice: $F(x)=-2\sin x \cos x=- \sin 2x+C.$ Then $\lim_{x \rightarrow} \frac{F(x)}{x}=\lim_{x \rightarrow \infty} \frac{-\sin 2x +C}{x} =0,$ Because $-1\le \sin x \le 1$.

Answer 2

Let $\bar{f}$ be the average of $f$.

Then $F(x) = F(0)+ \bar{f} x + \int_0^x (f(t)-\bar{f})dt$. It is not hard to see that $|\int_0^x (f(t)-\bar{f})dt| \le M$ for some $M$.

Then since ${F(x) \over x} = {F(0) \over x} + \bar{f} + { \int_0^x (f(t)-\bar{f})dt\over x}$ we have $\lim_{x \to \infty} {F(x) \over x} = \bar{f}$.

Answer 3

In general the limit is $\frac 1 {2\pi} (F(2\pi) -F(0))$. Hints for proving this: $a_k\equiv \int _{2k\pi}^{2(k+1)\pi} f(x)dx$ is independent of $k$ by periodicity. It follows that $\frac 1 n (a_1+a_2+..+a_n) = F(2\pi) -F(0)$ which gives $\frac {F(2\pi n)} n \to F(2\pi) -F(0)$. Thus $\frac {F(x)} x \to \frac 1 {2\pi} ( F(2\pi) -F(0))$ along the sequence $1,2,...$. Now use boundedness of $f$ to show that $\frac {f(x)} x \to \frac 1 {2\pi}(F(2\pi) -F(0))$.