Is the Epsilon-Delta definition of a limit not precise enough?

Question

Consider the function $f: \{-1, +1\}\to \mathbb R$ defined by

$f(x)= \arcsin (\frac{1+x^2}{2x})$.

Due to the following two inequalities :

(i) $1+x^2 \geq 2x$

(ii)$1+x^2 \geq -2x$ ,

the function can only be defined at $x=1$ and $x=-1$. I have learnt that the epsilon delta definition only includes those values of $x$ which are in the domain of $f(x)$. But in this case, the function isn't defined on either side of x=1.

So this is my question: Is it correct to say that the limit as $x$ approaches $1$ of $f(x$) is $\frac{\pi}{2}$ ?

Can the above question be given a definitive "yes" or "no" answer, or must it(unfortunately) vary from person to person?

If the latter, is the "precise" definition of a limit not precise enough?

How can the answer be proved or disproved using the epsilon delta definition?

I have also read that functions are by default continuous at isolated points. Can I conclude from the definition of continuity (the limit equals the value of the function evaluated at the point) that the limit must exist?

Note : Forgive my ignorance but I do not know a thing about topology. I'm looking for a detailed answer but in simple terms, preferably written in the language of calculus.

Thanks for the help.

Answer 1

The bottom line is:

Limits are only defined on limit points. Continuity is only defined on the domain.

If we have a point $p$ on the domain such that $p$ is a limit point, then continuity at $p$ is equivalent to the limit being the value. However, if we have a point not on the domain, continuity does not make sense (although we can enlarge the domain and define it at the point in order to force continuity, but I digress), and if we have a point on the domain which is not a limit point, then limit does not make sense.

The problem is explictly the following:

The definition of limit is: Let $x_0$ be a limit point of the domain. Then, we say $f(x) \stackrel{x \to x_0}{\to} L$ if for all $\epsilon>0$, there exists $\delta>0$ such that $0<|x-x_0|< \delta \implies |f(x)-L| < \epsilon$.

If $x_0$ is not a limit point, there exists a $\delta$ such that there is no $x$ with $|x-x_0|<\delta$. Therefore, any value of $L$ can be the limit. If we want to be really forceful and withstand the non-uniqueness of limit on metric spaces, then we can indeed let this be the case: every point on the codomain is a limit of $f(x)$ as $x \to x_0$ if $x_0$ is not a limit point. Then, for instance, it would hold that $f$ is continuous at $x_0$ if and only if $f(x) \stackrel{x \to x_0}{\to} f(x_0)$ even if $x_0$ is not a limit point, but that would be a very degenerate case.

What we conclude is that the case of $x_0$ not being a limit point results in a nuisance of non-uniqueness. Continuity is different:

The definition of continuity at $x_0$ is: $f$ is continuous at $x_0$ if for all $\epsilon>0$ there exists $\delta>0$ such that $|x-x_0|<\delta \implies |f(x)-f(x_0)| < \epsilon$.

There is no vacuous problem. Rather, the definition gives immediately that if $x_0$ is not a limit point, then $f$ is automatically continuous at $x_0$ (go through the definition and verify this). There is a certain degree of triviality, but no vacuity.

Summing up, your function $f$ is continuous, but limits on the points of the domain are not well-defined (at least, not unique when viewed under the usual definition).

PS: I acknowledge that there may be different definitions of continuity. However, the one used here is the special case of the definition used in topology, which is by far the most useful and well-behaved.

Answer 2

Usually when talking about limits of functions $f:D\subseteq \mathbb R \rightarrow \mathbb R$ in calculus, we work with functions whose domain is not a discrete set. Because of this, an extra condition on the limit is left out: the limit can only be taken at limit points of the domain. A point $c$ is a limit point of the domain $D$ if, for all $\delta > 0$, there exists $x\in D$ with $0 < |x-c| < \delta$. Note that $c \in D$ is not required. It should be fairly clear why this condition is desired in the definition of the limit. So the epsilon-delta definition of the limit at $c$ should be augmented with the condition $\forall(\delta > 0) \exists (x \in \mathrm{dom}[f]): 0 < |x-c| < \delta$.

The domain of your function $f$ is $\{-1, 1\}$. Like all discrete sets, it has no limit points, and so the limit is not defined anywhere.

Answer 3

Recall the (most commonly used) definition of $\lim\limits_{x\to 1} f(x) = \frac\pi2$:

$$\forall\epsilon > 0~\exists \delta>0: \forall x \in \operatorname{dom}(f): 0<|x-1|<\delta \implies |f(x)-\frac\pi2|<\epsilon$$

(Note that we need to exclude the situation $x=1$ to distinguish between approaching $1$ and being $1$ -- we are interested in the behaviour close to $x$.)

We see that if we choose $\delta < 2$, then there are no $x$ such that $x \in \operatorname{dom}(f)$ satisfying $0<|x-1|<\delta$. So the implication holds vacuously.

Therefore, we can say $\lim\limits_{x\to 1} f(x) = \frac\pi2$. Of course, this doesn't single out $\frac\pi2$; in fact, any value can thus be argued to be this limit. (For this reason, some sources would say that the limit does not exist.)

A similar situation holds for continuity. Depending on whom you talk to, they will either argue the function is (vacuously) continuous at $1$ or that its continuity is not well-defined at $1$.

Answer 4

It seems that $x$ is tacitly assumed to be real. In this case the given expression is defined only on the two-element set $S=\{-1,1\}\subset{\mathbb R}$. The set $S$ inherits from ${\mathbb R}$ the metric $d(x,y):=|x-y|$, and as $|1-(-1)|=2$ it follows that $S$ is a discrete metric (resp., topological) space. It follows that any function on $S$, in particular the function considered in the question, is continuous.

As for the limit $\lim_{x\to1} f(x)$ I think that one should not even consider such a limit. Formally, for each $\lambda\in{\mathbb R}$, and any given $\epsilon>0$, one (trivially) has $\bigl|f(x)-\lambda\bigr|<\epsilon$ as soon as $0<|x-1|<1$.

Answer 5

Given $\epsilon>0$, let us choose $\delta=2$.

If $|x-1|<\delta=2$, $x\in\{-1,1\}$, that is if $x=1$, then $$|f(x)-f(1)|=f(1)-f(1)=0<\epsilon$$ is satisfied. So, $\lim_{x\to1}f(x)=f(1)=\frac\pi 2.$