# justification of a limit

I encountered something interesting when trying to differentiate $F(x) = c$.

Consider: $\lim_{x→0}\frac0x$.

I understand that for any $x$, no matter how incredibly small, we will have $0$ as the quotient. But don't things change when one takes matters to infinitesimals? I.e. why is the function $\frac0x = f(x)$, not undefined at $x=0$?

I would appreciate a strong logical argument for why the limit stays at $0$.

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+1, I like this question. (I fear there might be duplicates, although I didn't find any.) Plus it's one of the few here to which I know the answer and no one answered yet. :) – JMCF125 Mar 8 '14 at 12:26
You may also benefit from "Why does factoring eliminate a hole in the limit?". I particularly like the "limits are about the journey" image in the answer from @Blue . – joeA Mar 8 '14 at 12:56

The limits are not about a value at a point, but about the values approaching that point.

I.e. why is the function $\frac0x = f(x)$, not undefined at $x=0$?

It is undefined at that point. However, its "neighbourhood" is defined, and that's what the limit gives you.

Thus $\lim_{x\to0}\frac0x=0$ means that $x$ approaching $0$ from both sides results in the same value, $0$, the limit must be defined from both sides and equal, i.e. the function is continuous in there, $\lim_{x\to0^-}\frac0x=\lim_{x\to0^+}\frac0x=\lim_{x\to0}\frac0x=0$.

## Edit:

This function ($f(x)=\frac0x$), alike $\frac1x$, is continuous in its domain, $\mathbb R\setminus\{0\}$. (it can't really be continuous for $x=0$, because $0$ is outside it's domain) And so I see the confusion, as $\lim_{x\to0^+}\frac1x=+\infty$. One could naïvely think this is because $\frac10=\infty$, but that is (if taken literally) outside the standard definition of the real numbers, and regarding limits, an abuse of notation for $\lim_{x\to0^+}\frac1x$. Unknowingly, one might have expected $\lim_{x\to0}\frac0x$ to be a thing of that sort ($\infty$, or at least undefinition of the limit), but as I said in the beginning, that's not what limits are about.

Besides what I have said, I was going to include something with the $(\varepsilon,\delta)$-definition of limits, but I don't know it very well and it is quite confusing. So, I'll follow the initial infinitesimal spirit of Calculus, now present in the more intuitive Non-standard analysis. We know $\forall x\in\mathbb R\setminus\{0\}, \frac0x=0$. Imagine $x$ is a very small number. No matter whether $x$ is positive or negative, and no matter how small, $\frac0x$ will be $0$. That means the stated property is extensible beyond $\mathbb R$. The limit is essentially another interpretation of that, denying numbers smaller than all reals by using numbers without a fixed value ($x\to0$ instead of fixed infinitesimal $x$). Sorry for not adding a formal theoretical basing to this yet, I'll try to do it ASAP.

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I think maybe I should talk more about continuity. I'll research and develop the answer, it still seems rather "dry" to me. – JMCF125 Mar 8 '14 at 12:42

Note that writing $f(x) = \frac0x$ results in $f(0) = \frac00$ wich is undefined. However, the singularity of $f$ is nice in the way that is can be continuously defined by $f(0) := 0$ (note the colon for defining the value). A limit is exactly this concept: What is the value of $f(x)$ when $x$ comes arbitrarily close to $0$, but not equal to $0$. The statement $$\lim_{x\to 0} \frac0x = 0$$ Means exactly that, and not, as you might think $\frac00 = 0$. The misconception is that $$\lim_{x\to x_0} \frac{f(x)}{g(x)} \neq \frac{\lim_{x\to x_0} f(x)}{\lim_{x\to x_0} g(x)}$$ For general $f,g$ (even if both are continuous!). This only works when $\lim_{x\to x_0} g(x) \neq 0$, wich is not the case in your example.

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You are in fact considering function $f: \mathbb R \setminus \{0\} \mapsto \mathbb R$ that is defined $f(x) = \frac 0x$ so it i equal $0$ on all its domain. Lets look at limit $\lim \limits_{x \rightarrow 0^+} f(x)$. Function is identically equal $0$ on every open interval $(0,\varepsilon)$ for $\varepsilon>0$. Hence right side limit is equal $0$. By analogy we also have $\lim \limits_{x \rightarrow 0^-} f(x)=0$. Whether there exists limit $\lim \limits_{x \rightarrow 0} f(x)$ is now a matter of convention. According to how I was taught it has no sense to talk about a limit of function outside its domain. But if you think otherwise, then since both one-sided limits exist and are equal $0$ the limit exists and is equal $0$.

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A limit of functions outside its domain is a very interesting concept, actually. This allows you to talk about $\lim_{x\to\infty} f(x)$ where ${\rm dom}(f) \subset \mathbb R \not\ni \infty$ – AlexR Mar 8 '14 at 12:37
Well, in my poit of view $\lim \limits_{x \rightarrow \infty} f(x)$ is improper limit, defined dfferently: $\forall \varepsilon >0 \exists N \in \mathbf R \forall x >N \; |a-f(x)| < \varepsilon$ – wroobell Mar 8 '14 at 12:40
You can define limits via neighborhoods and neighborhoods of $\infty$ are defined for $\mathbb R$ ;) But there are plenty of other limits "outside the domain". For example in the context of approximative units, such as the Poisson kernel. – AlexR Mar 8 '14 at 13:45

The reason is if you look at the function $f(x,y)=y/x$ then when you approach $(0,0)$ this should be invariant to the path you take in your case you walk along the path where $y=0$ (constant $0$) but when approaching over the path $y=x$ then it is constant $1$. So we say it is not defined.

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A function isn't just an expression, but you can think whether a single expression can be applied to an argument. The expression $0^{-1}$ is rather meaningless, so you don't know how to get the behavior of the function $f(x)=0\cdot x^{-1}$ at $x=0$ from the expression.

Limits are just a way to describe the behavior (if it looks consistent enough that the limit exists) around the point. It doesn't state anything about the value of the function at the point. That is if $a=\lim_{x\to b}f(x)$ then the function $f_1(x)=\left\{\begin{matrix}f(x)&\text{ if }x\neq b\\a&\text{ if }x=b\end{matrix}\right.$ is continuous.

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When we deal with limits we first see the result of the part after the sign $\lim$. In this case we have $\frac{0}{x}=0$. Than we have $\lim_{x\to0}{\frac{0}{x}}=\lim_{x\to0}0=0$.

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