Is it possible to simlify the follwing expression with a floor function:
$\lim_{n\to \infty}\frac{1}{n}\left(\left\lfloor \frac{n\tau}{T}\right\rfloor-1\right)$
Is it possible to simlify the follwing expression with a floor function:
$\lim_{n\to \infty}\frac{1}{n}\left(\left\lfloor \frac{n\tau}{T}\right\rfloor-1\right)$
For a real number $x$, $\lim_{n\to +\infty}\frac{\lfloor nx\rfloor}n=x$. Indeed, $nx-1\leq\lfloor nx\rfloor\leq nx,$ then we divide by $n$ and apply squeeze theorem.