0

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 $\epsilon$.

1 Answers1

0

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.

user803264
  • 376
  • 1
  • 3
  • What if the domain is undefined at a point which is not the point the limit is approaching? –  Jun 25 '20 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. – user803264 Jun 25 '20 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 $f$. – user803264 Jun 25 '20 at 17:32
  • https://www.quora.com/What-is-the-difference-between-mapping-into-versus-mapping-onto#:~:text=%E2%80%9CInto%E2%80%9D%20and%20%E2%80%9Conto%E2%80%9D,throughout%20mathematics%20in%20this%20way.&text=The%20function%20is%20a%20surjection,least%20one%20in%20such%20that%20. 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 . –  Jul 29 '20 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). –  Jul 29 '20 at 12:12
  • Shouldn't the mapping depend on the surjectivity,injectivity and bijectivity of the function rather then be a general rule however the statement $\forall x \in (c-\delta,c+\delta) \setminus {c} \cap \mathrm{dom}_f \ \exists y \in (L-\epsilon,L+\epsilon) \ s.t. \ y= f(x)$ be a general statement directly implied by the continuity of a function at $c$ that is synonymous with the limit existing at that point? –  Jul 29 '20 at 12:44