Periodic functions and Lyapunov exponents

Question

Suppose $f(x)$ is periodic. Or even quasiperiodic. Does it follow that $f(x)$ has a zero top Lyapunov exponent? I'm thinking that if we say $f(x)$ is periodic, then $f(x)=f(x+2\pi)$ as an example. This implies that $f$ is bounded and thus if we consider; $$\lambda_{\mathrm{top}}=\lim_{x\to\infty}\frac{1}{x}\ln|f(x)|=\lim_{x\to\infty}\frac{1}{x}\ln|f(x+2\pi)|$$ This gives; $$\lambda_{\mathrm{top}}=\lim_{x\to\infty}\left(\frac{1}{x}\ln|f(x)|-\frac{1}{x}\ln|f(x+2\pi)|\right)=\lim_{x\to\infty}(0)=0$$ Is this correct? and wouldn't this extend to quasiperiodic functions too?