Continuity of sin(1/x) on (0,1)

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In summary, The function sin(1/x) is continuous on the interval (0,1) by the definition of continuity, since it is defined at every point on the interval and the limit of the function as x approaches c exists and is equal to f(c) for all points except x=0, which is not in the interval. This is also true for the interval (0, infinite) as it is a subset of (0,1). Additionally, the composition of continuous functions (sinx and 1/x) is continuous on their common domain, further proving the continuity of sin(1/x) on (0,1). There is no need to use the epsilon-delta proof in this case.
  • #1
rsa58
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Homework Statement


how do you show that sin(1/x) is continuous on (0,1)? (i know it's also continuous on (0, infinite)).


Homework Equations





The Attempt at a Solution



|f(x)-f(xo)| = |sin(1/x)- sin(1/xo)|= |2sin((xo-x)\2)cos((xo+x)/2)|

=< 2|sin((xo-x)/(2xox))|=< |(xo-x)/(xox)|. is this inequality true? a similar one is used in a different example. if it is, why? is it because sin(x)=<x ? when x is positive? now since x<1 choosing [tex]\delta[/tex]=xo[tex]\epsilon[/tex] then if

|xo-x|<[tex]\delta[/tex] then |f(xo)-f(x)|<[tex]\epsilon[/tex]

is this answer correct? what about the endpoints?
 
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  • #2
i guess this is true for all xo and x different from 0 so it is true for (0, infinite) right?
 
  • #3
A cheap way would be to use the theorem that differentiability implies continuity
The derivative of [tex]\sin(\frac{1}{x})[/tex] is [tex]\frac{-1}{x^2}\cos(\frac{1}{x})[/tex].
The only place where the derivative is undefined is at x=0 which is not in the given interval so the function is continuous on the interval.

I don't really like that way because it seems out of order (differentiability comes after continuity)

Here's another.
Definition of Continuity: The function is defined at every point on the interval and
[tex]\lim_{x\to c}f(x) = f(c)[/tex]

We know that the function is defined at every point except 0, so it is defined in the interval.
The limit as x goes to c of the function is [tex]\sin(\frac{1}{c})[/tex] which is defined and equal to f(c) at every point except c=0, which isn't in the interval
Therefore, the function is continuous on (0,1)
 
  • #4
sinx is continuous, and 1/x is continuous on (0,1). The composition of continuous functions is continuous. So sin(1/x) is continouous on (0,1)

Are you required to prove that sinx itself is continuous?
 
  • #5
You say "(i know it's also continuous on (0, infinite))". If a function is continuous on a set, A, it is continuous on any subset of A.
 
  • #6
HallsofIvy said:
If a function is continuous on a set, A, it is continuous on any subset of A.

ANY subset of A? What if the subset has an empty interior? How can f inverse of an open subset of f(A) (in the subspace topology of R) be open in A?
 
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  • #7
What's the epsilon-delta proof for the continuity of sinx itself?
 
  • #8
mathboy said:
What's the epsilon-delta proof for the continuity of sinx itself?
One can prove that |sin(x) - sin(y)| <= |x - y|.
 
  • #9
mathboy said:
ANY subset of A? What if the subset has an empty interior? How can f inverse of an open subset of f(A) (in the subspace topology of R) be open in A?
The empty set and A itself are still open subsets of A.
 
  • #10
Ok, I got it. If A is given the subspace topology of R, then any continuous function restricted to A is still continuous.

But an epsilon-delta proof of this fact will not work if A is a set with an empty interior.
 
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  • #11
i see you have all posted great replies, but can anyone tell me if my proof actually works. i want to understand the proper use of the epsilon-delta proof. the teacher will surely ask us to do such a proof. it seems okay.
 
  • #12
There's no need to use the epsilon-delta proof, and I don't think your prof expects it.
 
  • #13
I don't think so either, the exercise does not seem to encourage you to take a look at the epsilon-delta kind of proofs.
 

1. What is the definition of continuity in mathematics?

The definition of continuity in mathematics is the property of a function where small changes in the input result in small changes in the output. In other words, a function is continuous if it has no abrupt changes or breaks in its graph.

2. How is continuity of a function determined?

To determine the continuity of a function, we need to check three conditions: 1) the function is defined at the point in question, 2) the limit of the function at that point exists, and 3) the limit and the value of the function at that point are equal.

3. What is the definition of sin(1/x)?

The function sin(1/x) is defined as the sine of the reciprocal of x. In other words, it is the sine of 1 divided by x.

4. Is sin(1/x) continuous on the interval (0,1)?

No, sin(1/x) is not continuous on the interval (0,1) because it is not defined at x=0. The function has a vertical asymptote at x=0, which means it has a break in its graph and does not satisfy the condition of continuity.

5. Can the continuity of sin(1/x) on (0,1) be fixed?

No, the continuity of sin(1/x) on (0,1) cannot be fixed because it is not possible to redefine the function at x=0 to make it continuous. The function will always have a vertical asymptote at x=0, which makes it discontinuous on the interval (0,1).

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