What's the boundary of $y = \sin (\frac{1}{x}), \; 0 < x \le 1$? I have figured out the graph of this function. It's oscillating around $x = 0$. But how can I define it's boundary then? I'm not sure about it.
The boundary of sin(1/x)
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calculus
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0It has no meaning to talk of the boundary of a function. You may intend the domain, the image, the graphics? You need a set to have a boundary. – 2012-09-30
1 Answers
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I'd suggest the segment of the y-axis from (-1,0) to (0,1) plus all the points on the graph itself - take any open disc about a point on said segment and theres got to be an oscillation of the function that passes through it.
Look up 'Topologist's Sine Curve' for more information.
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2That should be (0,-1) to (0,1). – 2012-09-30