Proving Continuity of a Piecewise Function

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Homework Help Overview

The discussion revolves around the continuity of a piecewise function defined as f(x) = x^2 for x ≤ 0 and f(x) = x + 1 for x > 0. Participants are tasked with determining the points of continuity and justifying their findings.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the continuity at x = 0, with some asserting that the function is not continuous there based on specific sequences. Others question the sufficiency of the textbook's example for proving continuity at points other than x = 0.

Discussion Status

The discussion is active, with participants providing insights and questioning the clarity of the textbook's explanations. Some suggest using known theorems about the continuity of polynomials for points away from x = 0, while others express confusion about the justification provided in the text.

Contextual Notes

Participants note that the textbook example may lack detail and clarity, particularly regarding the establishment of neighborhoods and the formal proof of continuity. There is also mention of the need to show limits for all sequences converging to x = 0.

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Homework Statement



Define
f = { x^2 if x \geq 0
x if x < 0​

At what points is the function f | \Re -> \Re continous? Justify your answer.

Homework Equations



A function f from D to R is continuous at x0 in D provided that whenever {xn} is a sequence in D that converges to x0, the image sequence {f(xn)} converges to f(x0).

The Attempt at a Solution



Check x=0. {1/n} converges to 0. {f(1/n)} converges to 1. f(0) = 0. 0 ≠ 1. Thus the function is not continuous at x=0.

When it comes to proving it's continuous elsewhere...that's where I have a problem.

The example in the book for proving that a function is continuous at every other point simply states that "If a sequence {xn} converges to x0, then there is an index N such that f(xn) = f(x0) for all indices n > N.
Thus, lim (n->∞) f(xn) = f(x0), and so f is continuous at the point x0."

How is that a proof? That logic doesn't even exclude x=0. Can someone either elucidate what they've done, or explain an alternate way of proving this function's continuity at all points other than x=0?

Thanks in advance.
 
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f(1/n) does not converge to 1. It converges to zero.
 
krissycokl said:

Homework Statement



Define
f = { x^2 if x \geq 0
x if x < 0​

At what points is the function f | \Re -> \Re continous? Justify your answer.

Homework Equations



A function f from D to R is continuous at x0 in D provided that whenever {xn} is a sequence in D that converges to x0, the image sequence {f(xn)} converges to f(x0).

The Attempt at a Solution



Check x=0. {1/n} converges to 0. {f(1/n)} converges to 1. f(0) = 0. 0 ≠ 1. Thus the function is not continuous at x=0.

As SammyS points out {1/n} → 0. But that doesn't settle the question anyway. You have to show the limit works for all sequences converging to 0 if you're going to prove it that way.
When it comes to proving it's continuous elsewhere...that's where I have a problem.

The example in the book for proving that a function is continuous at every other point simply states that "If a sequence {xn} converges to x0, then there is an index N such that f(xn) = f(x0) for all indices n > N."

No, I don't believe your book says that.

How is that a proof? That logic doesn't even exclude x=0. Can someone either elucidate what they've done, or explain an alternate way of proving this function's continuity at all points other than x=0?

Thanks in advance.

Do you have theorems such as "polynomials are continuous"? You can use that to show continuity for x away from 0. At 0 you need to show the right and left limits are both f(0).
 
Sorry, I copied the problem incorrectly. The piecewise function should read:

f(x) = x^2, if x<= 0
x+1, if x>0​

Thus, again:

Check x=0. {1/n} converges to 0. {f(1/n)} converges to 1. f(0) = 0. 0 ≠ 1. Thus the function is not continuous at x=0.

You have to show the limit works for all sequences converging to 0 if you're going to prove it that way.

This would be correct if I was trying to prove that it is continuous at x=0, but I'm showing that it is NOT continuous at x=0. Thus a counterexample is sufficient. What I'm having trouble with is discovering a general method of showing that ALL of a function is continuous (excluding the point x=0)

No, I don't believe your book says that.

My text is Fitzpatrick Advanced Calculus 2e, and I've scanned the page in question.

bogusproof.jpg


Do you have theorems such as "polynomials are continuous"? You can use that to show continuity for x away from 0. At 0 you need to show the right and left limits are both f(0).

No such theorem, at least not for a few sections. I think I understand, though. Using the sum and product properties of limits, I can say that for x<0, lim (n->∞) f(xn) = lim (n->∞) (xn2) = (lim (n->∞) x0))*(lim (n->∞) x0) = x02 = f(x0)

Then after repeating the same process for x>0, I would conclude that f(x) is continuous from (-∞,0)U(0,∞).

I suppose this is what that particular example in the book was trying to show, but it was woefully inadequate as an illustration of the process and seemed more general than it was meant to. I don't much like this textbook.
 
Last edited:
krissycokl said:
The example in the book for proving that a function is continuous at every other point simply states that "If a sequence {xn} converges to x0, then there is an index N such that f(xn) = f(x0) for all indices n > N.

LCKurtz said:
No, I don't believe your book says that.

krissycokl said:
bogusproof.jpg

The only reason that quote is correct is because the specific example to which it refers is constant nearby x if x ≠ 0. Very misleading to quote that "theorem" out of context.
 
I never claimed it was a theorem, and indeed referenced it as part of an example. But, yes, I can see where it was misleading.

My question stands, though. How is that sufficient justification, even within the example I've scanned? It seems to be glossing over important steps such as establishing a neighborhood of x for which the limit holds. Not to mention that my text doesn't establish the definition of a neighborhood until the next chapter.

Such a lax progression of "logic" seems out of place in what is supposed to be a formal proof, especially when it is the sole example the text offers order for proving continuity in an interval.

What I am asking is if someone can more clearly delineate those last few steps, because I do not feel this example sufficient for my understanding.

Thanks.
 
krissycokl said:
I never claimed it was a theorem, and indeed referenced it as part of an example. But, yes, I can see where it was misleading.

My question stands, though. How is that sufficient justification, even within the example I've scanned? It seems to be glossing over important steps such as establishing a neighborhood of x for which the limit holds. Not to mention that my text doesn't establish the definition of a neighborhood until the next chapter.

Such a lax progression of "logic" seems out of place in what is supposed to be a formal proof, especially when it is the sole example the text offers order for proving continuity in an interval.

What I am asking is if someone can more clearly delineate those last few steps, because I do not feel this example sufficient for my understanding.

Thanks.

I tend to agree with you that it isn't the best example and it isn't written in complete detail. It depends on how much detail is required. For example, he just asserts 1/n → 0 as n → ∞ without proof. It may be OK to assume that is obvious. Similarly, it may be OK to state the "non-theorem :smile:" is also obvious for this example given that the function is constant on those two intervals. If you give the author the benefit of the doubt on what is obvious, then his example is OK.

Normally in a problem like that, once you have covered the basics, you would assert continuity away from 0 since the formulas are polynomials, then show the left and right limits at zero are different, again using continuity of the left and right polynomials.

I don't know if I really addressed your question. But I do think you understand what is going on and wouldn't be too concerned about it.
 

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