# Epsilon-delta on a function with restricted range

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 $$\epsilon$$.

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.

• What if the domain is undefined at a point which is not the point the limit is approaching?
– user716881
Jun 25, 2020 at 17:24
• 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. Jun 25, 2020 at 17:30
• 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 $f$. Jun 25, 2020 at 17:32
• 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 .
– user716881
Jul 29, 2020 at 12:12
• 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).
– user716881
Jul 29, 2020 at 12:12