Definitions 10/12/17

In class we discussed the Sequential Criterion for Functional Limits, Divergence Criterion for functional limits, and Characterizations of continuity. We emphasized that we have built an understanding of functional limits and convergence that allow us to use several definitions interchangeably depending on the proof we are trying to complete.

Theorem 4.2.3 (Sequential Criterion for Functional Limits)

Given a function f\rightarrow \mathbb{R} and a limit point c of A, limf(x)=L iff for all sequences (x_n)\subseteq A satisfying x_n\neq c and (x_n)\rightarrow c, it follows that f(x_n)\rightarrow L.

Pf template:

Let \epsilon >0, and choose $latex\delat >0$ so that x\in (V_\delta (c) \setminus \{c\})\cap A.

\Rightarrow Start by assuming limf(x)=L and consider the arbitrary sequence (x_n)\subset A such that x_n \neq c and (x_n)\rightarrow c. Hence, we know that there exists V_\delta c with with the property that all x\in (V_\delta (c) \setminus \{c\})\cap A satisfy f(x)\in V_\epsilon (L). Notice that x_n is eventually in (V_\delta (c)\setminus \{c\} \cap A and thus f(x_n) is eventually in V_\epsilon (L).

Notice that this part of the proof uses topological definitions. The next part of the proof is a proof by contradiction.

\Leftarrow Now assume that lim_{x\rightarrow c} f(x)\neq L so that there exists an \epsilon_\circ >0 such that for all \delta>0 there exists an x\in (V_\delta (c) \setminus \{c\})\cap A such that f(x)\notin V_{\epsilon_\circ} (L). We choose \delta=\frac{1}{n} so that x\in (V_{\frac{1}{n}} (c) \setminus \{c\})\cap A but f(x_n)\notin V_{\epsilon_\circ} (L). Hence f(x_n) does not converge to L. Therefore when all sequences (x_n)\subseteq A satisfying x_n\neq c and (x_n)\rightarrow c, it follows that f(x_n) do not converge to L.

The Divergence Criterion for Functional Limits follows from this: If we can produce two sequences (x_n) and (y_n) in A with x_n\neq c and y_n\neq c and limx_n=limy_n=c, but limf(x_n)\neqlimf(y_n), then the functional limit lim_{x\rightarrow c} f(x) does not exist.

We can conclude the results in the Algebraic Limit theorem for Functional Limits by using the definition of functional limits and the Algebraic Limit Theorem for Sequences.

The Algebraic Limit Theorem:

Let f and g be functions defined on a domain A\subseteq \mathbb{R} and assume lim_{x\rightarrow c} f(x)=L and lim_{x\rightarrow c} g(x)=M. Then

  • lim_{x\rightarrow c} kf(x)=kL for all $latex k\in \mathbb{R}
  • lim_{x\rightarrow c} [f(x)+g(x)]=L+M
  • lim_{x\rightarrow c} [f(x)g(x)]=LM
  • lim_{x\rightarrow c} [f(x)/g(x)]=L/M provided M\neq 0.

Finally, the alternative definitions of continuity are stated in Theorem 4.3.2, the Characterizations of Continuity

Let f:A\rightarrow \mathbb{R} and let c\in A. The function f is continuous at c if any one of the following conditions hold:

  • For all \epsilon>0, there exists a \delta>0 such that |x-c|<\delta (and x\in A) implies that |f(x)-f(c)|<\epsilon.
  • For all V_\epsilon (f(c)), there exists an V_\delta (c) with the property that x\in V_\delta (c) (and x\in A) implies that f(x)\in V_\epsilon f(c).
  • For all (x_n)\rightarrow c (with x\in A), it follows that f(x_n)\rightarrow f(c).
  • If c\in L(A), lim_{x\rightarrow c} f(x)=f(c).

Posted on by Elaine Wissuchek
Categories: Definitions

One Response

  1. Jeremy LeCrone says:
    October 17, 2017 at 4:50 pm

    Thank you for the post Elaine.

    I will encourage you to review the conclusion to the second part of the proof of Theorem 4.2.3. You have the construction of a sequence (x_n) \subset A so that (x_n) \rightarrow c, and yet f(x_n) \notin V_{\varepsilon_0}(L). You’re conclusion here should be the contrapositive of the statement we are ultimately trying to prove. I would encourage you to write out the implication we have proven and see how to express the contrapositive of it, to confirm we have proven what we set out to prove…

    This would be a good exercise for this “Daily Definitions” blog, the remainder of the results you have listed are good to have in your notes, but not necessary for inclusion in these definition blogs…

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