Let $A = \bigl\{ t \sin (\frac {1}{t})\mid t \in (0, \frac{2}{\pi})\bigr\}$. Which of the following statement are true?
$1).$ $\sup(A) < \frac{2}{\pi} +\frac {1}{n\pi} $ for all $n \ge 1$
$2)$ $\inf(A) > \frac{-2}{3\pi} - \frac{1}{n\pi}$ for all $n \ge 1 $
$3)$ $\sup(A ) = 1$
$4)$ $\inf(A) = -1 $
My attempts:
I know that $-1 \le t \sin (\frac {1}{t}) \le 1$, so option 3 and option 4 is the correct answer.
Is it true or false?
On
The graph is a wave which dies off as $t$ approaches to $0$ from the right.
The supremum is the limit as $t$ approaches to the right end, $$t=2/\pi$$ and the supremum is $$2/ \pi $$
The infimum is attained at the local minimum closet to the right end which is at $t=0.2254769$ and the infimum is $y=-0.2172336$
Thus the options $(1)$ and $(2)$ are correct.
Since $$-1 \leq \sin(x) \leq 1$$ for any real $x$, it follows that $$-1 \leq \sin \Big(\dfrac{1}{x}\Big) \leq 1$$ for any real $x$, and it can be shown that $$-x \leq x\sin \Big(\dfrac{1}{x}\Big) \leq x$$ for any positive real $x$.
Since the values of $t$ are between $0$ and $\dfrac{2}{\pi}$, the values of $$t\sin \Big(\dfrac{1}{t}\Big)$$ must be between $-\dfrac{2}{\pi}$ and $\dfrac{2}{\pi}$, which are around $\pm 0.6366$ and thus the least upper bound and greatest lower bound could not be $1$ and $-1$, they have to be at least $\dfrac{2}{\pi}$ and $-\dfrac{2}{\pi}$.
Since $\sup(A) \leq \dfrac{2}{\pi}$, number 1 is clearly true as $\dfrac{1}{n\pi}$ is strictly positive.
Statement 2 however is false. The minimum value of $$t\sin \Big(\dfrac{1}{t}\Big)$$ is around $-0.21723$ which can be found using a method of numerical approximation, or some graphing software, I used desmos. The value of $-\dfrac{2}{3\pi}$ is around $-0.21221$ meaning that at some point the term $-\dfrac{1}{n\pi}$ would get small enough to make $-\dfrac{2}{3\pi} - \dfrac{1}{n\pi}$ as a whole greater than $\inf(A)$ which turned out to be just less than $-\dfrac{2}{3\pi}$. The value of $n$ which makes $-\dfrac{1}{n\pi}$ just big enough to make the statement $$\inf(A) > -\dfrac{2}{3\pi} - \dfrac{1}{n\pi}$$ false is $64$, meaning $$\inf(A) < -\dfrac{2}{3\pi} - \dfrac{1}{n\pi} \ \textrm{when} \ n \geq 64$$