Originally Posted by
Archie Meade Start with $\displaystyle sin(n{\pi})=0,\;\;\;n=0,\;1,\;2,\;3......$
$\displaystyle \displaystyle\ sin\left(\frac{(4n+1){\pi}}{2}\right)=1$
$\displaystyle \displaystyle\ sin\left(\frac{4n+3){\pi}}{2}\right)=-1$
Then if
$\displaystyle \displaystyle\ sin(n{\pi})=0\Rightarrow\ n{\pi}=\frac{1}{x}\Rightarrow\ x=\frac{1}{n{\pi}}$
shows that the graph crosses the x-axis at
$\displaystyle \displaystyle\ x=\frac{1}{\pi},\;\;\frac{1}{2{\pi}},\;\;\frac{1}{ 3{\pi}},\;\;,\frac{1}{4{\pi}},.....$
an "infinite" number of times approaching x=0.
The function is non-periodic and approaches 0 as x approaches $\displaystyle \infty$