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Find the least upper bound (LUB) and greatest lower bound (GLB) of $\{x\sin(1/x):x>0\}$

My Attempt: Since limit of given function is $0$ as $\lim_{x\to f(x)}g(x) = 0$ when $f(x)\to0$ and $g(x)$ is bounded. So $\operatorname{LUB}(f) = \operatorname{GLB}(f) = 0$ But it don't resemble with actual answer. Please help me.

K.defaoite
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Ankit
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    "LUB" is the least upper bound of the function over the set of interest, i.e. the $\sup$. You are confusing it with the limsup as $x\to 0$. Similar for "GLB". –  May 25 '21 at 04:46
  • Yes. Thank you. I got LUB = $2/π$ and GLB = $-2/3π$. Am I right ? If yes then I'm wrong with actual answer. One of the answer is GLB < $-2/3π$ – Ankit May 25 '21 at 05:43
  • @Gae. S , Please see my comment. I'm right or wrong. – Ankit May 25 '21 at 06:07
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    Also consider the behavior of the function as $x \to \infty$. – Paul Sinclair May 25 '21 at 16:42
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    I do not see where you got the LUB and GLB values. The local extrema of this function do not occur at nice neat fractions of $\pi$ or their inverses. One trick that might be helpful is to substitute $t = \frac 1x$ as the variable. The resultant function of $t$ will have the same extrema, though at different locations. – Paul Sinclair May 25 '21 at 17:16
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    Please use correct MathJax to make your maths look pretty. For example, use \{...\} for the set you are defining, and $\lim_{x\to ???}f(x)g(x)$ renders as $\lim_{x\to ???}f(x)g(x)$ (I would have edited it in, but I am struggling to work out some of your meanings). – user1729 May 26 '21 at 09:23

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