Discontinuity: Asymptotic discontinuity

[samir]

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.

 

[Greg]

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

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

Can a function have a vertical asymptote but no horizontal asymptote?

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

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

That's correct.

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.

 

[Deveno]

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."

 

[I like Serena]

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)

 

[Deveno]

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.