How Do You Find All Boundary Points for S={(x,y); 0 < x< 1 and y=sin(1/(1-x))}?

[tara123]

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 don't understand how I can actually find all the points.. can anyone get me started??

The help is appreciated!

 

[Office_Shredder]

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

 

[Hurkyl]

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?

 

[Steve154]

Hey! Thats funny I have the same question in my practice questions too!
Anyone know how to do it? My prof gives no examples!

 

[Hurkyl]

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?

 

[Steve154]

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

 

[tara123]

haha yah don't feel bad I am pretty lost too!

 

[tara123]

Above ur taking the new function of y to be sin(x) right? .. this seems wrong lol no offence

 

[Steve154]

k me and tara123 clearly don't know what were talking about can anyone help us?

 

[Office_Shredder]

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?

 

[HallsofIvy]

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).

 

[aeluik]

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.

 

[tara123]

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?

 

[tara123]

sorry i meant sin(1)=PI/2

but just one other thing, I've bin thinking and y=sin(1/1-x) o<x<1 0 isn't even contained in the set...so how is it a boundary point if 0 is strictly less than x?
Help!