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On James Stewart's Calculus Early transcendental it says:

The definition of limit says that if any small interval $(L - \epsilon , L + \epsilon)$ is given around $L$, then we can find an interval $(a - \delta, a + \delta)$ around a such that $f$ maps all the points in $(a - \delta, a + \delta)$ (except possibly a) into the interval $(L - \epsilon , L + \epsilon)$.

However, ''small'' is not specific which contradicts the notion of a formal definition.

In the definition it says $\forall \epsilon$ but functions with a restricted range e.g. $\sin(x)$ it is impossible for $f$ to map all the points in $(a - \delta, a + \delta)$ (except possibly a) onto the interval $(L - \epsilon , L + \epsilon)$, for all $\epsilon$.

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The definition requires $(a - \delta, a + \delta)$ to map into $(L - \varepsilon, L + \varepsilon)$, not onto it. There's no requirement for all of $(L - \varepsilon, L + \varepsilon)$ to be covered; we simply want the image of every point within $\delta$ of $a$, to be within $\varepsilon$ of $L$.

As a concrete example (simpler than $\sin$), consider the constant function $f(x) = 1$. Let's also take $a = 2$. We show that it approaches the limit $L = 1$ as $x \to 2$. For any given $\varepsilon > 0$, I'll choose $\delta = 10$. Then, $$x \in (a - \delta, a + \delta) = (-8, 12) \implies f(x) = 1 \in (L - \varepsilon, L + \varepsilon).$$ This proves that $\lim_{x \to 2} f(x) = 1$. Note that not every point in $(L - \varepsilon, L + \varepsilon)$ is in the range of $f$, but more importantly, the points around $a$ map within the interval. This is in the spirit of continuity: we need points nearby $a$ to map near to $L$; we don't really care if they take the full tour.

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  • $\begingroup$ What if the domain is undefined at a point which is not the point the limit is approaching? $\endgroup$
    – user716881
    Jun 25, 2020 at 17:24
  • $\begingroup$ That's a more nuanced question, with two answers: one for beginners, and one for the more advanced. For the beginners, normally limits don't use $x \in (a - \delta, a + \delta)$, but instead require a punctured neighbourhood $(a - \delta, a + \delta) \setminus \{a\}$. This is usually expressed by $0 < |x - a| < \delta$. In this way, we can talk about limits to points not in the domain in the function, so long as the function is defined at every point around $a$. If it's defined only to the left or right, we have the concepts of left and right limits to fill in the gaps. $\endgroup$
    – user803264
    Jun 25, 2020 at 17:30
  • $\begingroup$ For the more advanced, we tend to define limits at accumulation points of the function's domain, i.e. where $(a - \delta, a + \delta) \setminus \{a\}$ will intersect with the function's domain, no matter how small $\delta$ becomes. For example, if a function's domain was $\{1/n : n \in \Bbb{N}\}$, then the only point where limits would be sensible would be $0$ (even though there is no interval around $0$ contained the domain). We replace $x \in (a - \delta, a + \delta)$ with $x \in (a - \delta, a + \delta) \cap \operatorname{dom} f$, i.e. only consider $x$ in the domain of $f$. $\endgroup$
    – user803264
    Jun 25, 2020 at 17:32
  • $\begingroup$ quora.com/…. says that ''into'' means injection which would suggest that $\forall x_1,x_2 \in (c−\delta,c+\delta)\setminus{c}\cap \mathrm{dom}_f $ if $f(x_1)=f(x_2)$ then x1=x2 . $\endgroup$
    – user716881
    Jul 29, 2020 at 12:12
  • $\begingroup$ however this does not make sense say for $f(x) = 1$, I choose 2 distinct points $a_1,a_2 \in (c-\delta,c+\delta)\setminus \{c\} \cap \mathrm{dom}_f$ then $f(a_1)=f(a_2)=1$ however $a_1 \neq a_2$ by definition thus the function is not injective but it is still (obviously) continuous so shouldn't it be a map onto (surjection) not into (injection). $\endgroup$
    – user716881
    Jul 29, 2020 at 12:12

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