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Does it make sense to calculate a limit of a function at some point $a$, if there is no neighbourhood of $a$ on which $f$ would be defined? For example $$\lim_{x\to 0}x\ln{x}$$ doesn't exist, because there is no $\delta>0$ s.t. $(-\delta,\delta)\subset\mathcal{D}_f$. So, by definition, if we wanted to verify, whether the implication $$x\in(-\delta,\delta)\Rightarrow f(x)\in(-\epsilon,\epsilon)$$ hold, this would make no sense, because for $x\in(-\delta,0\rangle$, there "is no value of $f$".

On the other hand, it would make sense if we talked about the right-sided limit, that is $$\lim_{x\to 0^+}x\ln{x} $$ now, this exists and equals $0$.

Also, we have a nice fancy theorem saying, that for $a\in\mathbb{R}$ the limit $$\lim_{x\to a}f(x)$$ exists if and only if both \begin{align*} \lim_{x\to a^-}f(x)\\ \lim_{x\to a^+}f(x) \end{align*} exist and equal the same number $L\in\overline{\mathbb{R}}$, where $\overline{\mathbb{R}}=\mathbb{R}\cup\{\infty,-\infty\}$.

For the case of $f(x)=x\ln{x}$, this makes no sense, right? as $x\rightarrow 0^-$, there is no limit, because for any left $\delta$-neighbourhood of $0$, it doesn't even make sense to talk about $f(x)$, whenever $x\in (-\delta,0)$.

Question: Should we, as a very first thing, in a definition of a limit of $f$ at $a$, require that $f$ is defined on some both-sided neighbourhood of $a$?

Michal Dvořák
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2 Answers2

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Recall the fundamental and formal definition of a limit :

Let $ f $ be a real-valued function defined on a subset $D_f \subseteq \mathbb R$. Now, let $x_0$ be a limit point of $D_f$ and let $L$ be a real number. We say that $$\lim_{x \to x_0} f(x) = L$$ if for every $\varepsilon > 0$ there exists a $ \delta>0$ such that, for all $x\in D_f$, if $0<|x-x_0|<\delta $, then $ |f(x)-L|<\epsilon$.

Note : A limit point is number $x_0$ such that for all $\varepsilon>0$, there exists a member of the set $y\neq x$ such that $|y-x_0|<\varepsilon$.

Rebellos
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  • Wait, but then $(0,\infty)$ has limit point $0$ (because, for any $\epsilon$, take $y=\epsilon / 2$, then set of $y\neq x$ s.t. $|y-x|<\epsilon$ (as you described it) is always non-empty. Now from the condition that $x\in D_f$, we automatically take the ones from the $(0,\delta)$ anyways, don't we? So it actually DOES make sense... uhm? – Michal Dvořák Dec 14 '18 at 23:08
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    @MichalDvořák Letting an element of your set take the arbitrary values of your $\varepsilon-\delta$ definition is a bad practice, since in any case that element may not even belong for such $y$. Generally, if you consider a space $X$, then the absolute value becomes a norm $|\cdot|X$ and then $x \in X$ isn't supposed to be a real number anyway, thus no such substitution will lead anywhere, even if you bypassed the above. Recall also that for $\lim{x\to x_0} f(x)$ to exist, its side limit must exist and be equal, meaning : $\lim_{x\to x_0^-} = \lim_{x\to x_0^+}f(x) = L \in \mathbb R$. – Rebellos Dec 14 '18 at 23:12
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Some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $0$ but, according the more general definition of limit, we can write that

$$\lim_{x\to 0} x\ln x =\lim_{x\to 0^+} x\ln x= 0$$

that is we don't need to specify $x>0$ since the points $x\le0$ are out of the domain of definition for the given function.

That important issue has been deeply discussed here

and notably to that specific answer

Note that not all sources must be considered at the same level. Using the first definition is a reasonable way to deal with limits at a lower (high school) level whereas, in a more advanced context, the second more general definition should be adopted.

Moreover, note that, according to the first definition, the discussion of limits would reduce, as in the present case, to a (more or less) trivial determination on the domain of definition for functions which is of course a very different topic.

Therefore, when we deal with limits, I strongly suggest to refer to the more general definition excluding points outside the domain of definition for the function considered.

user
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