"Proving" |x| is continuous on all points: Is my understanding sufficient?

[Integrate]

The definition of |x| is as follows

[math]|x| = \begin{cases} x &\text{if }x \ge 0 \\ -x &\text{if }x < 0 \end{cases}[/math]We also know the following:
Every polynomial function is continuous everywhere on (−∞, ∞)

Meaning that both x and -x of our piecewise function are continuous on all points.


For further rigor the transition point of x=0 fits all three criteria for a limit to be continuous at a point, f(a) must be defined, the limit as x goes to a must exist, and the must they must equal each other.


Is this sufficient enough to "prove" that |x| is continuous at all points?


James Stewart 7th edition Ch. 2.5 exercise 68(a)

 

[Deleted member 4993]

The definition of |x| is as follows

[math]|x| = \begin{cases} x &\text{if }x \ge 0 \\ -x &\text{if }x < 0 \end{cases}[/math]We also know the following:
Every polynomial function is continuous everywhere on (−∞, ∞)

Meaning that both x and -x of our piecewise function are continuous on all points.


For further rigor the transition point of x=0 fits all three criteria for a limit to be continuous at a point, f(a) must be defined, the limit as x goes to a must exist, and the must they must equal each other.


Is this sufficient enough to "prove" that |x| is continuous at all points?


James Stewart 7th edition Ch. 2.5 exercise 68(a)

What is the definition of continuity according to James Stewart 7th edition - the book you are following?

 

[JeffM]

I am a little suspicious of that proof

[math]- \infty < x < 0 \implies |x| = - x \implies |x| \text { is continuous on } (- \infty, \ 0) \\ \text {because, as a polynomial, } - x \text { is continuous on any open interval.}[/math]
That looks fine to me.

[math]0 < x < \infty \implies |x| = x \implies |x| \text { is continuous on } (0, \ \infty) \\ \text {because, as a polynomial, } x \text { is continuous on any open interval.}[/math]
That too looks fine to me.

But it seems to me that, rather than merely assert a conclusion about continuity at x = 0, you need to show explicitly

[math]|0| \in \mathbb R \text { and, for arbitrary real number } \epsilon > 0, \\ \exists \ \delta \text { such that } \delta < x < 0 \implies |- x - |0|| < \epsilon \\ \text {and } 0 < \delta < x \implies |x - |0|| < \epsilon.[/math]
Proofs about continuity at a point relate to a point in an open interval.

 

[BigBeachBanana]

I believe this is your book's definition. For [imath]x=0[/imath], I would show [imath]\lim_{x\to 0^-}|x|=\lim_{x\to 0^+}|x|=f(0)[/imath].

 

[Integrate]

I am a little suspicious of that proof

[math]- \infty < x < 0 \implies |x| = - x \implies |x| \text { is continuous on } (- \infty, \ 0) \\ \text {because, as a polynomial, } - x \text { is continuous on any open interval.}[/math]
That looks fine to me.

[math]0 < x < \infty \implies |x| = x \implies |x| \text { is continuous on } (0, \ \infty) \\ \text {because, as a polynomial, } x \text { is continuous on any open interval.}[/math]
That too looks fine to me.

But it seems to me that, rather than merely assert a conclusion about continuity at x = 0, you need to show explicitly

[math]|0| \in \mathbb R \text { and, for arbitrary real number } \epsilon > 0, \\ \exists \ \delta \text { such that } \delta < x < 0 \implies |- x - |0|| < \epsilon \\ \text {and } 0 < \delta < x \implies |x - |0|| < \epsilon.[/math]
Proofs about continuity at a point relate to a point in an open interval.

This is why I put proof in quotations. I knew an actual proof would need a delta epsilon proof to complete it. I am just wondering if my word explanation is sufficient to show an understanding.

Here is the book question.

 

[Integrate]

What is the definition of continuity according to James Stewart 7th edition - the book you are following?

As BigBeachBanana has provided me

Which is how I describe it in my original post.

I would also say that both sides of a limit must equal the same value and be defined at that value.

 

[BigBeachBanana]

As BigBeachBanana has provided me View attachment 32670

Which is how I describe it in my original post.

I would also say that both sides of a limit must equal the same value and be defined at that value.

JeffM has showed you the case [imath]x<0[/imath] and [imath]x>0[/imath].
For [imath]x=0[/imath], I think what you said is sufficient to show that f is continuous everywhere.
Since [imath]\lim_{x\to 0^-}|x|=\lim_{x\to 0^+}|x|=0 \implies \lim_{x\to 0}|x|=0[/imath].
Further more, [imath]f(0)=0[/imath].
By the definition of continuity, [imath]\lim_{x\to 0}|x|=f(0)=0[/imath]. The function is continous at 0.

 

[Steven G]

There are three things to show that a function f(x) is continuous at x=a.
1a) \(\displaystyle \lim_{x\rightarrow a^-}f(x)\ exists\)
1b) \(\displaystyle \lim_{x\rightarrow a^+} f(x)\ exists\)
2) \(\displaystyle \lim_{x\rightarrow a^-}f(x)\ = \lim_{x\rightarrow a^+}f(x)\ = L\)
3) f(a) = L

Yes, polynomials are continuous everywhere on its domain. Since f(x) changes definition at x=0, it is possible that f is discontinuous there.
Use my definition above to show whether or not f(x) is continuous at x=a.

 

[Integrate]

There are three things to show that a function f(x) is continuous at x=a.
1a) \(\displaystyle \lim_{x\rightarrow a^-}f(x)\ exists\)
1b) \(\displaystyle \lim_{x\rightarrow a^+} f(x)\ exists\)
2) \(\displaystyle \lim_{x\rightarrow a^-}f(x)\ = \lim_{x\rightarrow a^+}f(x)\ = L\)
3) f(a) = L

Yes, polynomials are continuous everywhere on its domain. Since f(x) changes definition at x=0, it is possible that f is discontinuous there.
Use my definition above to show whether or not f(x) is continuous at x=a.

I did this. So does that my explanation is sufficient to answer the question from my textbook?

 

[BigBeachBanana]

I did this. So does that my explanation is sufficient to answer the question from my textbook?

It depends on how rigorous you're required to be, but I studied calculus from the same textbook and saw their solutions. I'd say that it's adequate to their standard and also this is not Real Analysis.

 

[Steven G]

I did this. So does that my explanation is sufficient to answer the question from my textbook?

For further rigor the transition point of x=0 fits all three criteria for a limit to be continuous at a point, f(a) must be defined, the limit as x goes to a must exist, and the must they must equal each other.

The lines above shows that you know what needs to be shown, which is good. The problem is that you only stated what you need to show.
You failed to state what the left hand and right hand limits equals as x approaches 0 and you failed to state what f(0) equals.
I guess that I need to be harsh and say that what you wrote is pure nonsense.

 

[Steven G]

It depends on how rigorous you're required to be, but I studied calculus from the same textbook and saw their solutions. I'd say that it's adequate to their standard and also this is not Real Analysis.

BBB,
I agree that this is not Real Analysis but I also agree that just knowing what needs to be shown and showing it are not the same.
Sure, at the calculus 1 level one get away with just answering the requirements of a function to be continuous at a point, but just stating the requirements really should not be sufficient.