MHB Discontinuity: Asymptotic discontinuity

  • Thread starter Thread starter samir
  • Start date Start date
  • Tags Tags
    Discontinuity
AI Thread Summary
Asymptotic discontinuity refers to points where a function is discontinuous due to limits being infinite, leading to vertical and horizontal asymptotes. The function f(x) = 1/x exemplifies this, as the limit approaches infinity when x nears zero, indicating a vertical asymptote. It is clarified that while infinity is not a number, the limit notation is well-defined in mathematical contexts. The discussion also explores whether functions can have vertical asymptotes without horizontal ones, and vice versa, confirming that such scenarios are possible. Overall, asymptotic discontinuity plays a crucial role in understanding the behavior of functions near points of discontinuity.
samir
Messages
27
Reaction score
0
Hi again!

I thought I would finish off my previous posts on discontinuity by discussing asymptotic discontinuity. So let's focus on these alone in this thread.

I'm not familiar with the origin of the term "asymptote", but from what I can tell, it has asymptotic discontinuity to thank for for its existence. That is to say asymptotic discontinuity is the reason we have these vertical and horizontal lines that we call asymptotes.

So what is asymptotic discontinuity? The simplest explanation I can think of is that it's the point at which a function is discontinuous because the limit at the given point is unreachable, because it's infinite!

Assume we have the following function.

$$f(x)=\frac{1}{x}$$

How do we show that this function has vertical and horizontal asymptotes? My current understanding is that we have to show that the limit is infinite. But how do we do that?

For the vertical asymptote, I believe that we can write something like this:

$$\lim_{{x}\to{0^{\pm}}}f(x) \rightarrow \frac{1}{x}= \infty$$

Letting x approach 0 implies that the quotient approaches infinity.

But I don't think we can write this:

$$\lim_{{x}\to{0^{\pm}}}f(x)= \infty$$

Because x is not allowed to be equal to zero. The function is undefined at x=0.

$$\lim_{{x}\to{0^{\pm}}}f(x)= \text{undefined}$$

Or am I looking at this the wrong way? Is the limit in fact equal to infinity? Surely nothing can really be equal to infinity as infinity is not even a number?

What about the horizontal asymptote? How do we show that this function has a vertical asymptote? Do we have to solve for the dependent variable and kind of do the same thing in reverse?

Now, is it only rational functions that have asymptotes? I have mostly seen this type of function having an asymptote. I know tangent function also has asymptotes. But that has to do with the definition of what a tangent function is. We can't divide by zero.

Can a function have a vertical asymptote but no horizontal asymptote? I would say yes.

Can a function have a vertical asymptote and a horizontal asymptote? I would say yes to that too.

But can a function have a horizontal asymptote alone, and no vertical asymptote? I'm not so sure of that.
 
Last edited:
Mathematics news on Phys.org
samir said:
The simplest explanation I can think of is that it's the point at which a function is discontinuous because the limit at the given point is unreachable, because it's infinite!

That's true for vertical asymptotes and they occur as the function approaches some constant value. For horizontal asymptotes the limit is some constant, for example

$$\lim_{x\to\infty}\tan^{-1}(x)=\dfrac{\pi}{2}$$

samir said:
Can a function have a vertical asymptote but no horizontal asymptote?

Yes. $\tan(x)$ is an example.

samir said:
Can a function have a vertical asymptote and a horizontal asymptote? I would say yes to that too.

That's correct.

samir said:
But can a function have a horizontal asymptote alone, and no vertical asymptote? I'm not so sure of that.

Yes, it can. $\tan^{-1}(x)$ is such a function.

***

$\lim_{x\to0^\pm}\dfrac1x=\pm\infty$ and it's purely a notational convention. As you say, infinity is not a number and can't be treated as such in the sense of real numbers. There are systems, such as the extended reals, where infinities are treated somewhat differently.
 
A bit of humor:

"Infinity isn't a number, it's a place."

-"What place is that?"

"Not anywhere near here."

The notion $\lim\limits_{x \to 0^+} \dfrac{1}{x} = \infty$ is actually well-defined, and can be stated without mentioning infinity:

"For any positive (real) number $M$ there exists a positive real number $\delta$ such that if $0 < x < \delta$, then $\dfrac{1}{x} > M$".

Informally, "$\dfrac{1}{x}$ grows without bound as $x$ nears 0 from the right."
 
greg1313 said:
There are systems, such as the extended reals, where infinities are treated somewhat differently.

I especially like the sentence:
With these definitions $$\overline{\mathbb{R}}$$ is not a field, nor a ring, and not even a group or semigroup. However, it still has several convenient properties.


In other words, it violates all known algebraic structures, but it's still useful. (Wink)
 
I like Serena said:
I especially like the sentence:
With these definitions $$\overline{\mathbb{R}}$$ is not a field, nor a ring, and not even a group or semigroup. However, it still has several convenient properties.


In other words, it violates all known algebraic structures, but it's still useful. (Wink)


One can also form the *hyperreals*, which not only contain "infinities", but also *still* remain a field.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...

Similar threads

Replies
20
Views
2K
Replies
3
Views
1K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
8
Views
5K
Replies
6
Views
1K
Back
Top