- #1

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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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- Thread starter tara123
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- #1

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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!

- #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=x

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

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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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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