# Are these definitions of limits the same?

So I’m just beginning to teach myself real analysis and I came across the definition of a limit point: for real numbers, A is a limit point of a subset of real numbers U if every deleted neighborhood of A contains a point in U. This made me think of limits like the limits used in calculus and I can see how it inspired the name limit point and vice versa. So I decided to attempt to write out (guess) what I thought the definition of a limit of function, in calculus, would be before I read it.

What seemed straight forwards to me, a new reader in the subject: “the limit as x goes to c of $$f(x)$$ is L if(f) for every deleted neighborhood of c, N, L is a limit point of $$\{f(x)|x\in N\}$$.”

The definition in what I was reading: “The limit as goes to c of f(x) is L if f is defined on some deleted neighborhood of c and, for every $$\epsilon>0$$ there is a $$\delta>0$$ such that $$|f(x)-L|<\epsilon$$ if $$0<|x-c|<\delta$$.”

Are these equivalent? My gut intuition is that they are but I know real analysis has a reputation for producing unintuitive results.

Obviously there are differences; I would not have used distance (the subtraction and absolute value), the idea of “narrowing in” by matching deltas with epsilons, etc. Is there something more subtle I’m missing? If they are the same, why is one approach/definition used and not the other? Is it for pratical or just historical reasons?

Edit: After reading some answers, would it work if I changed a to the?

• Saying that '$L$ is a limit point of $\{f(x):\ x\in N\}$' is weaker than the condition of $f$ having $L$ as limit as $x\to c$. This is because of the a. It allows for other limit points. For example, $f(x)=\begin{cases}x,&1\geq0\\-1,&x<0\end{cases}$ satisfies the property with $c=0$ and $L=1$. It also satisfies the property with $L=-1$ and the same $c$. – MoonLightSyzygy Jan 10 at 15:11
• Note that just replacing a by the would turn the condition into one that is way too strong. In most cases the set $\{f(x): x\in N\}$ will have many limit points, even when $f$ has limit at $c$. However, if the limit $L$ exists, then it must be the only limit point that belongs to all sets $\{f(x): x\in N\}$. This condition is not sufficient, though. – MoonLightSyzygy Jan 10 at 15:16
• Above I wanted $f(x)=1$ for $x\geq0$. – MoonLightSyzygy Jan 10 at 15:23
• @MoonLightSyzygy Thanks. I did not realize the a vs the. – Benjamin Thoburn Jan 10 at 20:16
• The definition of limit can be stated in terms of neighborhoods. If for every neighborhood $B$ of $L$ there exists a corresponding deleted neighborhood $A$ of $a$ such that $f(A\cap D) \subseteq B$ then $f(x) \to L$ as $x\to a$ ($D$ is domain of $f$). Your definition misses the key aspect that for all values of $x$ sufficiently near $a$, $f(x)$ should be near $L$. – Paramanand Singh Jan 16 at 2:22

No. Consider $$f(x)=\sin(1/x)$$ with the origin added, near $$x=0$$.
If $$\{f(x)|x\in N\}$$ has several limit points, your definition falls apart.
• Would an example of this be anything that’s not a function, something like $y^2=x$? I’m not sure though because this doesn’t have a limit by the standard definition either. – Benjamin Thoburn Jan 10 at 17:30