Is the given function continuous at x= 0? f(x) = {sin(1/x) if x≠0, 1

[MidgetDwarf]

Homework Statement

Prove or disprove that f(x) = { sin (1/x) if x≠0, 1 if x =0} is continuous on [0,infinity] by definition of continuity or one of its equivalent definitions.

Relevant Equations

Definition of continuity:

Let f: D → R^q. Let a∈D. We say that f is continuous at a if every neighborhood V of f(a) ∃ a neighborhood U of a (which depends on V) such that if x∈D∩U, then f(x)∈V. (Note that neighborhood of a point in X^n was defined to be a set that contains an open set containing the point x).

The equivalent definitions:

(1)If ∀ε>0, ∃δ>0 such that if ||x-a|| < δ, then ||f(x) - f(a)|| < ∈. (note ||. || is defined to be the standard metric on R^n).

(2) If (Xn) is any sequence of elements of D which converges to a, then the sequence (f(xn)) converges to f(a).

Discontinuity is also defined by (2), ie., showing that the negation of (2) holds, or

The function f is not continuous on a∈D iff there is a sequence (Xn) in D which converges to a, but (f(Xn)) does not converge to f(a).

I know that the function, g(x)= sin(1/x) has infinite oscillations when the values of x get closer and closer to 0. So its limit does not exist (from graphing it). However, the way that we defined f(x), at x=0, f(x)=1, but f(x)= sin(1/x) on (0,infinity).

I have an issue in general showing that piecewise defined functions are continuous/not continuous at a point without resorting to a limit argument.

I am assuming that the way how the function is defined, it fixes the discontinuity issue of g(x) at x=0. ?
I want to show |f(x) - f(0)| <ε, whenever |x-0|<δ.

I know that sin(1/x ≤ |1| ∀x∈R.

Busy work:

So |f(x)- f(o)| = |sin(1/x) - f(0)|≤ |sin(1/x)| + 1 ≤ 1+1=2

but |f(x) -f(o)|= |1 -1|=o.

So my confusion about continuity regarding piecewise defined functions is preventing me from going forward.

 

[Mark44]

So |f(x)- f(o)| = |sin(1/x) - f(0)|≤ |sin(1/x)| + 1 ≤ 1+1=2

but |f(x) -f(o)|= |1 -1|=o.

The last equation isn't true -- you're essentially saying that f(x) = 1 for any x. Notice that for some values of $\epsilon$ there is no $\delta > 0$ for which $|x - 0| < \delta \Rightarrow |f(x) - f(0)| < \epsilon$

 

[MidgetDwarf]

The last equation isn't true -- you're essentially saying that f(x) = 1 for any x. Notice that for some values of ϵ there is no δ>0 for which |x−0|<δ⇒|f(x)−f(0)|<ϵ

Hmm. I have to think more about your statement. I know that f(x)=1when x=0. So if I pick ε very small (near 0), then the second equation is true. But it is not always the case? I will think about your comment some more. Analysis is not that straightforward for me.

Ahh. I noticed an error in my problem statement. The domain is [0, infinity). We want to prove/disprove that f is continuous at the point 0, not that it is continuous on [0,infinity). So would my previous work still not be true in this case?

I tried an easier problem to practice a bit.

g(x) = { xsin (1/x) if x≠0, 1 if x =0} is continuous at x=0, where the domain of g is [0, infinity) , is this continuous. Yes/No?

I am assuming no.

Proof. Note that {Xn} = 1/n is defined on [0,infinity) , and converges to 0,
but |g(Xn) - g(0)|= |(1/n)(sin(n) - 1|≤ |1/n - 1 |, but g(Xn) does not converge to 1. Therefore, g is not continuous at 0.

Does this work?

 

[MidgetDwarf]

I think I have the proof. I will post tomorrow. Thank you Mark for pointing out the error.

 

[Mark44]

Hmm. I have to think more about your statement. I know that f(x)=1when x=0. So if I pick ε very small (near 0), then the second equation is true. But it is not always the case? I will think about your comment some more. Analysis is not that straightforward for me.

$\epsilon$ doesn't even need to be all that small -- take $\epsilon = .5$.
Can you find a number $\delta$ so that for all x within $\delta$ of 0, then sin(1/x) will be within 1/2 of 1? Think about things in terms of the graph of f(x) = sin(1/x).