High School What is a general definition of a limit?

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A limit is defined in the context of extended real numbers, particularly when considering cases where x approaches infinity. The equivalence of limits at infinity and finite limits hinges on the continuity of the function involved. Specifically, for a sequence converging to a limit, the definition requires that for any ε>0, there exists an N such that all terms beyond N are within the ε-neighborhood of the limit. In contrast, limits involving infinity are treated differently due to the absence of neighborhoods around infinite values. Understanding these distinctions is crucial for correctly applying the concept of limits in mathematical analysis.
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I was given this definiton of a limit:

Let us have a function ##f: \mathbb{R} \rightarrow \mathbb{C}##. Let ##x_0 \in \mathbb{R}^*## and ##L \in \mathbb{R}^*##. We say ##L## is a limit of a function ##f## for ##x## goes to ##x_0## if

##\forall \varepsilon >0 \, \exists \delta >0 \, \forall x \in P_{\delta}(x_0): f(x) \in U_{\varepsilon}(L)##

(##P## and ##U## are neighborhoods)

Is this definition valid for every type of limit?
I suppose that it is because we are in extended real numbers. But the definition of a limit when ##x_0 = \infty## and let's say ##L=\infty## is different. Why are these definitions equivalent? Isn't the key that ##U_{\varepsilon}(\infty)=\left(\frac {1}{\varepsilon},\infty\right)##?
 
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What does "I was given" mean?

That definition defines a continuous function ##f,## not a limit. What both connects them is
$$
\lim_{n \to \infty} f(x_n)=f(\lim_{n \to \infty}x_n)=f(x_0)
$$
so the limits both exist, i.e. are finite, and if ##\mathbf f## is continuous. This can be used as an equivalent definition of a at ##x_0## continuous function, but you need the definition of a limit first.

It doesn't make much sense for infinities since they have no neighborhoods. That's why converges to a finite number and grows beyond all finite numbers are treated differently.

A limit ##x_0## of a sequence ##(x_n)_{n\in\mathbb{N}}## is:
$$
\forall \, \varepsilon> 0\,\exists \,N(\varepsilon)\in \mathbb{N}\, \forall\, n>N(\varepsilon)\, : \, x_n\in U_\varepsilon(x_0).
$$
and in case ##x_0=\infty ##
$$
\forall\,M\in\mathbb{R}\,\exists\, N(M)\in \mathbb{N} \, \forall \, n>N(M)\,: \, x_n >M.
$$
and likewise for ##x_0=-\infty .##
 
Last edited:
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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