Question on Boundary Points

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Question on Boundary Points!!

Determine all boundary points:

S={(x,y); 0 < x< 1 and y= sin(1/(1-x)}


I'm really confused! I dont understand how I can actually find all the points.. can anyone get me started??

The help is appreciated!
 

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  • #2
Office_Shredder
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S is going to be a curve, so it looks like it's asking you to find what's the boundary at x=0 and at x=1. At x=1 of course you're going to run into a snag because 1/(1-x) doesn't act so nice there, so start by finding the boundary point at x=0
 
  • #3
Hurkyl
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Maybe start somewhere eaiser -- can you find any boundary points at all? Can you find all boundary points that lie in some easily-described subset of the plane?
 
  • #4
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Hey! Thats funny I have the same question in my practice questions too!
Anyone know how to do it? My prof gives no examples!
 
  • #5
Hurkyl
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You could try starting with an easier problem -- can you find any boundary points at all? Can you find all boundary points that lie in some easily-described subset of the plane?
 
  • #6
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well sin(x) would probably be easier but how to find the actual points. if u take x=0 then sin (0)=0 and sin(1)=π/2
 
  • #7
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haha yah dont feel bad im pretty lost too!
 
  • #8
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Above ur taking the new function of y to be sin(x) right? .. this seems wrong lol no offence
 
  • #9
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k me and tara123 clearly dont know what were talking about can anyone help us???
 
  • #10
Office_Shredder
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So for example, if you had S={(x,y)|y=x2+1 0<x<1} you can see that the boundary points of that curve are going to be exactly at (0,1) and (1,2). Now why is that. For any e>0, you can find a point on the curve S (x,y) such that |(x,y)-(0,1)|<e and also for any e>0, you can find a point on the curve (x,y) such that |(x,y)-(1,2)|<e. It's pretty easy here because y=x2+1 is a continuous function, so if you're told all the points x>0 are included, then the value at x=0 must be included.

So for the set S that you have, it's continuous away from x=1. So by the same process, we conclude that it has a boundary point at x=0. You should be able to find that point by yourself.

The tricky part is for x=1. It's not continuous there. What's the behavior of sin(1/(1-x)) as x approaches 1? What values of (1,y) might be part of the boundary?
 
  • #11
HallsofIvy
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First, have you drawn a sketch of the set?
S={(x,y); 0 < x< 1 and y= sin(1/(1-x)} is essentially the graph of y= sin(1/(1-x)) for 0< x< 1. (One thing I would be inclined to do is let u= 1-s so this is y= sin(1/u) for 0< 1< u.) Look closely at what happens around u= 0 (x= 1).
 
  • #12
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360px-Sin1over_x.svg.png

I found an image for when y=sin(1/u), but when you look at u=0, there is never a defined value for it as the function doesn't exist at u=0. I don't see how that would apply to our function.
 
  • #13
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yah so for x=0 y=sin(1)=pi
and for x=1 i just look at what the graph does as it approaches 0?
 
  • #14
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sorry i meant sin(1)=PI/2

but just one other thing, ive bin thinking and y=sin(1/1-x) o<x<1 0 isnt even contained in the set...so how is it a boundary point if 0 is strictly less than x???
Help!
 

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