# 0 divided by 0

Not necessarily. The use of L'Hopital's rule is essentially the application of a Taylor expansion for the numerator and denominator functions and considering the first non-zero term. Some smooth non-zero functions can have Taylor expansions where each and every coefficient is zero. For example $$f(x) = e^{-\frac{1}{x^{2}}}$$ at x=0 has Taylor coefficients of 0 to all orders, yet clearly if I consider g(x) = 2f(x) then the ratio $$\frac{f(x)}{g(x)}$$ at x=0 is simply 1/2, despite L'Hopital's rule being inapplicable yet the functions are both infinitely differentiable at x=0.

You should be careful, this is a bad counter example, you are subconsciously simplifying the fraction by g(x). g(x) does not exist in x=0.
The example you gave above is actually a classical example where the limit is undefined (not 1/2).

My point was that the function is infinitely differentiable (and thus continuous) everywhere, including 0, yet it is not equal to its Taylor expansion at x=0. Thus the application of L'Hopitals rule is invalid for said function at x=0. However, the limit $$\lim_{x \to 0} \frac{f(x)}{2f(x)}$$ is valid because the function is well defined as usual for any $$x \neq 0$$ and the whole point of using limits is that you don't plug in the value directly.

The reason I brought it up is to illustrate that although things like the derivatives used in L'hopital's rule are limits there's a difference between a ratio of limits and a limit of ratios for certain pathological functions and Magneto's attempt to be patronising was ill founded (like every time he opens his mouth).

The example you gave above is actually a classical example where the limit is undefined (not 1/2).
This is utterly wrong. It is obvious that

$$\lim_{x\rightarrow 0} \frac{\exp(-1/x^2)}{2\exp(-1/x^2)} = \frac{1}{2} .$$

Good work!

This is utterly wrong. It is obvious that

$$\lim_{x\rightarrow 0} \frac{\exp(-1/x^2)}{2\exp(-1/x^2)} = \frac{1}{2} .$$

Good work!

You too, what in "g(x) does not exist in x=0" did you fail to understand? Back to stalking, eh?

Why do you think the denominator or numerator in the said fraction need be defined at the limit point? Do you have similar problems with

$$\lim_{x\rightarrow 0} \frac{ 1/x}{2/x} ?$$

Are you still confused, or are we about to see yet another "Tach makes elementary errors and refuses to accept correction" saga?

You too, what in "g(x) does not exist in x=0" did you fail to understand? Back to stalking, eh?
If you convert the limit notation into the standard epsilon-delta notation of analysis you'll see that that doesn't matter. It is stated explicitly here that said definition of limit doesn't depend on the value of f at the point in question, as $$0 < x-p < \delta$$ is built into the definition, not allowing for $$0 = x-p$$.

Why do you think the denominator or numerator in the said fraction need be defined at the limit point? Do you have similar problems with

$$\lim_{x\rightarrow 0} \frac{ 1/x}{2/x} ?$$

Nope , both examples are functions that are identically equal to 1/2 everywhere except x=0 where they are not defined. That's all.If you have difficulty understanding this and you need to write all the possible combinations of $$\frac{g(x)}{2g(x)}$$ where g(x) is not defined for x=0, please continue.

Nope , both examples are functions that are identically equal to 1/2 everywhere except x=0 where they are not defined. That's all.If you have difficulty understanding this and you need to write all the possible combinations of $$\frac{g(x)}{2g(x)}$$ where g(x) is not defined for x=0, please continue.
Guest254 said:
Why do you think the denominator or numerator in the said fraction need be defined at the limit point?
In fact, if you find it simpler, you might like to try to justify that

$$\lim_{x\rightarrow 0} \frac{\exp(-1/x^2)}{2\exp(-1/x^2)}$$

does not exist, as per your claim.

My point was that the function is infinitely differentiable (and thus continuous) everywhere, including 0, yet it is not equal to its Taylor expansion at x=0.

agreed.

Thus the application of L'Hopitals rule is invalid for said function at x=0.

agreed.

However, the limit $$\lim_{x \to 0} \frac{f(x)}{2f(x)}$$ is valid because the function is well defined as usual for any $$x \neq 0$$ and the whole point of using limits is that you don't plug in the value directly.

Agreed. I thought that you were simplifying by $$g(x)$$ and I pointed out that $$g(x)$$ is not defined in x=0, so you cannot simplify. In effect you have defined a class of functions ($$\frac{g(x)}{2g(x)}=\frac{1}{2}$$) everywhere except x=0. If you want to find the limit, you can do it with the epsilon - delta definition or you can use an arbitrary sequence $$x_n->0$$ but not l'Hospital.

The reason I brought it up is to illustrate that although things like the derivatives used in L'Hospital's rule are limits there's a difference between a ratio of limits and a limit of ratios for certain pathological functions and Magneto's attempt to be patronising was ill founded (like every time he opens his mouth).

agreed.

both examples are functions that are identically equal to 1/2 everywhere except x=0 where they are not defined.
Which is irrelevant to the definition of the limit, as I just pointed out.

Consider a related issue, you have a function f on some open interval D = (a,b). On D let f be infinitely differentiable. f(a) is meaningless because $$a \not\in D$$, likewise for f(b). However, $$\lim_{x \to a_{+}}f(x)$$ (ie from above) and $$\lim_{x \to b_{-}}f(x)$$ (ie from below) do exist because you only evaluate f on some element of D.

This is also covered on the page I linked to where it comments that even when p is not in the set you're considering you can still define a one sided limit. If you can define the limit from both sides then you have a well defined notion of the limit of the function, else it is one sided. It is thus possible to even have a well defined two sided limit of say $$\lim_{x \to 0}f(x)$$ even if f is defined only on $$D = [-a,a]\backslash \{0\}$$ for a>0. In fact that is precisely what the $$f(x) = e^{-\frac{1}{x^{2}}}$$ example is, it has a well defined limit at x=0 because both one sided limits agree, even though its value at x=0 needs to be separately specified.

If you disagree please demonstrate the problem with something more than just words, let's see you show a contradiction using the epsilon-delta notation. It's not particularly elaborate notation, I'm just asking you to rise to the level of that Wiki page. Guest, I imagine, would ask for something even more detailed but he seems to like his analysis more than I.

If you disagree please demonstrate the problem with something more than just words, let's see you show a contradiction using the epsilon-delta notation.

Can't you read? I just pointed out that you can find the limit with either the epsilon - delta method or with arbitrary sequences. What you cannot do is to simplify by $$g(x)$$ and this is what it appeared that you were doing. So, you are not doing that, you clarified your position in the subsequent posts.

Can't you read? I just pointed out that you can find the limit with either the epsilon - delta method or with arbitrary sequences.
Why do you do this to yourself? How many times will someone need to quote your error before you admit to it?

You have claimed the limit does not exist. This is utterly false.

Look, it's not a big deal - no one would have guessed you had a grounding in analysis, and you've done nothing to change that position. All you're currently doing is cementing your reputation as a bit of a Wally. I say again - good work!

Why do you do this to yourself? How many times will someone need to quote your error before you admit to it?

You have claimed the limit does not exist. This is utterly false.

The second part of the post was a mistake, now can you stop trolling?

Hurrah - Tach admits to error in elementary subject! It only took 12 posts this time!

Hurrah - Tach admits to error in elementary subject! It only took 12 posts this time!

I am glad I made your day, from post 22 to 29 I only count 8 posts, not clear how you counted 12. So why do you continue trolling then?

Last edited:
You don't think my corrections of your claims are post worthy?

You don't think my corrections of your claims are post worthy?

Not really.

Well I guess that's to be expected!

Holy shoot i had no idea that this thread would be so popular!