The given function of x can be factorized as follows:
f(x)=(sinx)² / x·cosx=(sinx / x)·(sinx / cosx)=(sinx / x)·tanx Therefore, we have:
lim[x→½π]f(x)=lim[x→½π](sinx / x)·tanx=lim[x→½π](sinx / x)·lim[x→½π]tanx=2/π·lim[x→½π]tanx
At x=½π, the function tanx is discontinuous. (or if x→½π-0 ⇒ tanx→positive infinity, x→½π+0 ⇒ tanx→negative infinity)
Therefore, the limiting value of the given function does not exist.