Why the curve r(t) approaches a circle as t approaches infinity

[eclayj]

Both statements 1 and 2 are given as an explanation of why the original statement is true, but I don't understand why you can use statement 2 (since in the original vector equation you do not have Sin²(t), -Cos²(t))

Show why r(t) = <e^-t, Sin(t), -Cos(t)> approaches a circle as t →∞.

1. As t→∞, x(t)→0

2. And as t→∞, (y(t))²+(z(t))²=1

What I do not see, intuitively, is why the second statement is used and how it shows that r(t) = <e^-t, Sin(t), -Cos(t)> approaches a circle as t →∞.

I keep thinking that as t→∞, Sin(t) and -Cos(t) do not exist (the functions oscillate). And I don't understand how 2., (y(t))²+(z(t))²=1, relates beck to the original vector equation, Sin(t), -Cos(t)

 

[Office_Shredder]

Try this simpler question. Explain why r(t) = <0, sin(t), -cos(t)> is a circle. If you can do that you're 90% of the way to answering this problem.