[tex]\frac{\pi}{2}[/tex]
Step-by-step explanation:
When [tex]n->[/tex]0 ,[tex]\frac{Sin(kn)}{n}->k[/tex]
This is because,
using l'hospital's rule,[tex]Lt_{n->0}\frac{sin(kn)}{n}=\frac{kcos(n)}{1} =\frac{k\times 1}{1} =k[/tex]
In the given question,[tex]k=\frac{n\times\pi}{2}[/tex]
So,the given limit evaluates to [tex]\frac{n\pi}{2}[/tex]
