Question

I have a question about uniform continuity in advanced calculus.

Answer 1

I have the problem where I need to prove sin(x)/x is uniformly continuous on (0,1). The solution manual states that since sin(x)/x goes to 0 as x goes to 0, we have uniform continuity on (0,1). It states the theorem " If I is a closed bounded interval and f is continuous on f:I -> R then f is unifomly continuous on I" I dont get the connection.....

Answer 2

that means, from a larger POV, since the sin x/ x is continuous from neg infinity to infinity, then it's continuous over 0,1

Answer 3

? its not defined at 0. how can it be continuous at 0?

Answer 4

and we are talking uniform continuity ( not the same as continuity )

Answer 5

that was an example.
note that the brackets are ( not [ so 0 is not included.

Answer 6

yes but to show its uniform continuous on (0,1) the theorem states that we need to show its continuous on closed bounded I = [0,1] , it says that showing the limit is 0 as x goes to 0 proves this theorem. how can we be continuous on [0,1] when we are undefined at 0?

Answer 7

well it does not say i = [0,1] but its says its a closed bounded interval [a,b]

Answer 8

so the brackets are [ not ( ?
they are quite different is describing Real sets.

Answer 9

um, we are dealing with both. Do you know what uniform continuity on a set E vs continuous on a set E is?

Answer 10

unifrom continuity iff all epsilon > 0 , some delta > 0 s.t. |x-a|, where x,a elemnts of R, implies |f(x) - f(a)| < epsilon

Answer 11

the real question is, can we call a function f continuous on [a,b] if f(a) does not exist, but the limit as x approaches a from the right does?

Answer 12

strangely... f(0) does exist... cept it's because of the limit as x-->0 from the right and left.

Answer 13

f(0) is undefined ...

Answer 14

but the limits from both sides(neg inf and pos inf are defined, right?

Answer 15

I figured it out...