Find supremum of the set

Question

Let $A=\left\{t~\sin(\frac{1}{t}): t\in (0,\frac{2}{\pi})\right\}$. Then what would be suprimum of $A$.

It looks like supremum will occur at $\frac{2}{\pi}$ and supremum value will be $\frac{2}{\pi}$.

Answer 1

The supremum is indeed $\frac{2}{\pi}$. It cannot be more than $\frac{2}{\pi}$, because $|t| \leq \frac{2}{\pi}$ and $|\sin(\frac{1}{t})| \leq 1$.

Since $t \sin(\frac{1}{t})$ is continuous and $\frac{2}{\pi}\sin(\frac{1}{\frac{2}{\pi}}) = \frac{2}{\pi}\sin(\frac{\pi}{2}) = \frac{2}{\pi}$, it holds that $\lim_{t \rightarrow \frac{\pi}{2}} \left( t \sin(\frac{1}{t})\right) = \frac{2}{\pi}$, and $\frac{2}{\pi}$ is the supremum.