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On p96, chapter 5 in Spivak (3th edition), the definition of limit is given:

The function $f$ approaches the limit $l$ near $a$ means: for every $\epsilon >0$, there is some $\delta > 0$, such that for all $x$, if $0 < |x-a| < \delta$, then $|f(x)-l| < \epsilon$

Later in the same chapter (p104), it is mentioned that there must be $\delta > 0$ such that $f(x)$ is defined for $x$ satisfying $0 < |x-a| < \delta$, otherwise the clause $0 < |x-a| < \delta$, then $|f(x)-l| < \epsilon$ wouldn't make sense.

Shouldn't this remark be part of Spivak's definition of limit?

So the definition should better be:

Let $f$ be a real valued function, at least defined in some neighborhood of $a$ (not necessarily containing $a$). Then,the function $f$ approaches the limit $l$ near $a$ means: for every $\epsilon >0$, there is some $\delta > 0$, such that for all $x$, if $0 < |x-a| < \delta$, then $|f(x)-l| < \epsilon$

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    $\begingroup$ Nah, you can still consider limits when the function is defined in a set that only accumulates in $a$. $\endgroup$ – orole Feb 5 '18 at 13:45
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If you want to get technical, the definition makes sense in the following more general situation: say that $f$ is defined on a set $E \subset \mathbb R$, and you want to define the limit of $f$ at some $a \in \mathbb R$. Then you want $a$ to satisfy the following property : for all $\epsilon>0$, the set $(a-\epsilon, a+\epsilon) \cap E$ is not empty.

(the set of $a$ satisfying this property is called the closure of $E$).

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