Need a bit of help with a Calculus problem.

[Ampera]

What is the limit of F(X) = 1/X^2 (X Squared) as X approaches 0.

Does it exist, and why?

My answer is that it does not exist, reason being a limit can not be an infinite number, and that furthermore it has to be a real number, which infinity is not.

I however have been told be alternate sources that it does exist and that limits can be infinite.

Thanks.

[Andy Watson]

Try putting  \$x=0\$

And why do think that answer can't be \$\infty\$ ?

[Brumby]

That's not a calculus problem. Calculus deals with limits where delta x approches zero. You are dealing with a function that is discontinuous at X=0 which is a different thing altogether.

 

[Ampera]

That's not a calculus problem. Calculus deals with limits where delta x approches zero. You are dealing with a function that is discontinuous at X=0 which is a different thing altogether.

It was in a calculus class. Shoot me for not knowing.

[Ampera]

Try putting  \$x=0\$

And why do think that answer can't be \$\infty\$ ?

Because that would become an undefined answer due to division by zero.

Am I correct in that this has no limit? And is my reasoning correct?

[iaeen]

I think this is more of a philosophical debate than a calculus problem. Whether or not we agree that infinity is an "acceptable" solution to the limit, theres no arguing the fact that as X goes to zero, the function does indeed approach positive infinity.

I lean towards calling infinity the solution rather than "DNE" since this is distinct from something like \$\lim_{x \to 0}\frac{|x|}{x}\$, which truly does not exist since it approaches 1 from the right and -1 from the left.

Edit: having a go at better math formatting.

[Brumby]

In my book, the limit of that function for the condition given is infinity.

My view on limiting conditions is that the 'X' value (in this case) is so close to the limit that it is indistinguishable from the limiting value.

This follows on from the understanding I remember from high school calculus in regards to integration ... that the area under a curve is the sum of an infinite number of infinitely thin slices.  The width of these slices cannot be zero width - or the area of each and, thus, the sum of them would be zero .... but they are infinitely small.

[Ampera]

Thanks for the help, My final conclusion is that it's technically both in that, while the limit does not technically exist, answering it as infinity (or negative infinity) is a nicer way of showing how and why it does not exist.

[Andy Watson]

My final conclusion is that it's technically both in that, while the limit does not technically exist, answering it as infinity (or negative infinity) is a nicer way of showing how and why it does not exist.

Be careful. I would say that \$\lim\limits_{x \to 0}\$ of \$\frac{1}{x^2}\$ is \$+ \infty\$, but \$\lim\limits_{x \to 0}\$ of \$\frac{1}{x}\$ depends which way you approach it.

[IanB]

I think it might be instructive to compare it with the limit of 1/x as x goes to 0.

With 1/x^2 the limit is the same as x approaches from both the left and the right. So there is no ambiguity about the answer.

With 1/x you can either approach positive infinity or negative infinity according to which side you start. So what exactly should the limit be in that case? Somehow it is not clear whether it should be +inf, -inf, or some other value in between.

This, perhaps, is the subtext of the question.

[iaeen]

Actually, the more I think about it, the more I disagree with your assertion that the solution must be a real number.

Why wouldn't the concept of a limit translate onto (for example) the complex plane?

When you get passed introductory undergrad math classes, useless technicalities like this get thrown out the window. On the other hand, that is when math really starts to mess with your head 

[iaeen]

My final conclusion is that it's technically both in that, while the limit does not technically exist, answering it as infinity (or negative infinity) is a nicer way of showing how and why it does not exist.

Be careful. I would say that \$\lim_{x \to 0}\$ of \$\frac{1}{x^2}\$ is \$+ \infty\$, but \$\lim_{x \to 0}\$ of \$\frac{1}{x}\$ depends which way you approach it.

Is that LaTeX code?!?

I had no idea the forum could interpret that. Neato!

[Andy Watson]

When you get passed introductory undergrad math classes, useless technicalities like this get thrown out the window. On the other hand, that is when math really starts to mess with your head 

If you're working towards becoming an engineer don't worry about whether it exists or not, just treat it as a useful concept or tool - as you would with imaginary numbers. Let the mathematicians get bogged down in the philosophy of it

[Andy Watson]

Is that LaTeX code?!?

Yes. Use backslash dollar to delimit inline equations, or dollar dollar for stand-alone equations.

[iaeen]

When you get passed introductory undergrad math classes, useless technicalities like this get thrown out the window. On the other hand, that is when math really starts to mess with your head 

If you're working towards becoming an engineer don't worry about whether it exists or not, just treat it as a useful concept or tool - as you would with imaginary numbers. Let the mathematicians get bogged down in the philosophy of it

Oh, I'm already an engineer, but at the university I minored in math as well as spent 9 semesters on the math department payroll as a calculus tutor. I've lived in both worlds 

I've also actually formally studied philosophy too 

[Ampera]

Yea, this has been much help. Thanks.

[CatalinaWOW]

There is actually a lot of engineering value in discontinuous functions.  And as you get further into math the tools exist to deal with them formally.  These tools deal with different kinds of infinity, and these find application in classifying the time required to solve different types of problems and in things like cryptography.

Just keep your mind open, and keep biting off small chunks  that you can deal with.

[T3sl4co1l]

Whatever anyone else might will it to be, the fact remains: the correct answer is \$ + \infty \$.

The limit exists and is equal from both sides.

Infinity isn't a proper number, but that's fine for this situation.

It is not undefined!

Tim

[bson]

My final conclusion is that it's technically both in that, while the limit does not technically exist, answering it as infinity (or negative infinity) is a nicer way of showing how and why it does not exist.

Be careful. I would say that \$\lim_{x \to 0}\$ of \$\frac{1}{x^2}\$ is \$+ \infty\$, but \$\lim_{x \to 0}\$ of \$\frac{1}{x}\$ depends which way you approach it.

Is that LaTeX code?!?

I had no idea the forum could interpret that. Neato!

Yep, it's MathJax.  https://www.mathjax.org/

[bson]

It is not undefined!

This.  x^2 is always positive.  If it were 1/x however it would be undefined since it would be \$-\infty\$ when approached from the negative.

[helius]

This type of problem is usually considered under the rubric of Real Analysis, which in the US is normally taught prior to Calculus (some districts call it "pre-Calculus").
\$ \lim\limits_{x\to 0}{1\over x^2} \$ doesn't converge to any finite value: but it increases without limit to \$ \infty \$ from above and below, at the vertical asymptote \$ x=0 \$. Formulas with denominators have vertical asymptotes at the zeros of their denominator part: \$ x^2 \$ has a single zero at \$ x=0 \$, so that is where \$ {1\over x^2} \$ has its vertical asymptote.
It's true that "infinity is not a number", but for convenience it's used in limit formulas. Some formulas have different limits from above and below, or no limit at all (for example, \$ \lim\limits_{x\to 0}{1\over x^3} \$ goes to \$ \infty \$ from above, but to \$ -\infty \$ from below; and \$ \lim\limits_{x\to\infty}{\mathrm{sin} x} \$ doesn't converge at all).

[T3sl4co1l]

and \$ \lim\limits_{x\to\infty}{\mathrm{sin} x} \$ doesn't converge at all).

FYI: try \sin
\$ \sin \theta \$


Tim

[hans]

One sided limits are expressed like:

\$ \lim\limits_{x\to 0_{ }^{+}}{1\over x} = +\infty \$
\$ \lim\limits_{x\to 0_{ }^{-}}{1\over x} = -\infty \$

A two-sided limit implies that both left and right hand side limits are equal. Then the function is continuous, differentiable, etc. at x=0 So above example does not have a limit at x=0. But for this:

\$ \lim\limits_{x\to 0_{ }^{+}}{1\over x^2} = \infty \$ = \$ \lim\limits_{x\to 0_{ }^{-}}{1\over x^2} = \infty \$ = \$ \lim\limits_{x\to 0}{1\over x^2} = \infty \$

The limit does exist.
Things get even more fun in multi variable calculus, because you can approach a point from any direction.

If a function does not have a two-sided limit at a certain point, lots of normal Calculus falls apart and you need to be extra careful applying it near that function value, because there are jumps or gaps in the function.

For example: differentiate the Heaviside (unit step) function. You get the impulse function, or better known as the Dirac delta function. Although the latter is called a function; it is actually a distribution. And unfortunately, that theory was skipped in my calculus, signals & systems courses.

Nevertheless, discontinuities and one-sided limits are seen quite a lot in signal theory like Laplace of Fourier, because most of these transformations are unilateral. Otherwise the table of transformations would not converge for minus infinity. This poses problems if you have a causal system that has an initial value at t=0⁺.

Another typical case is the fundamental theorem of Fourier that describes what the value of a signal is represented by it's Fourier coefficients near discontinuities (namely the average of the left and right side limit).

[JacquesBBB]

Your problem is a problem of definition. The definition of a finite limit is not the same as the definition of an infinite limit.

- First thing to realise : \$ f(x) = {1\over x^2} \$ is not defined for \$ x=0 \$ .

But it is not necessary for a function to be defined on a point to have a limit on that point. The definition of limit is a way to extend the function in places where it is not defined.

One says that \$ f(x) \longrightarrow +\infty \$  when \$ x \longrightarrow 0 \$ if for any  \$ A > 0 \$, there exist \$ \epsilon \$ such that
\$
\vert x \vert  < \epsilon \implies  f(x) > A
\$

You can easily demonstrate that this is the case for \$ f(x) = 1/x^2 \$

Indeed, if \$ \vert x \vert  < 1\sqrt{A} \$, then  \$ f(x) > A \$.

So it is perfectly valid to write  \$ \lim_{x \longrightarrow 0} {1\over x^2} = +\infty \$.

[pitagoras]

It is very important that you understand deeply JacquesBBB answer.