# Advanced Calculus : Theory and Practice by John Srdjan Petrovic

By John Srdjan Petrovic

Sequences and Their Limits Computing the LimitsDefinition of the restrict houses of Limits Monotone Sequences The quantity e Cauchy Sequences restrict more suitable and restrict Inferior Computing the Limits-Part II genuine Numbers The Axioms of the Set R outcomes of the Completeness Axiom Bolzano-Weierstrass Theorem a few concepts approximately RContinuity Computing Limits of services A overview of services non-stop capabilities: ARead more...

summary: Sequences and Their Limits Computing the LimitsDefinition of the restrict houses of Limits Monotone Sequences The quantity e Cauchy Sequences restrict more desirable and restrict Inferior Computing the Limits-Part II genuine Numbers The Axioms of the Set R effects of the Completeness Axiom Bolzano-Weierstrass Theorem a few options approximately RContinuity Computing Limits of features A evaluate of features non-stop features: a geometrical standpoint Limits of features different Limits homes of continuing capabilities The Continuity of easy services Uniform Continuity homes of continuing capabilities

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**Extra resources for Advanced Calculus : Theory and Practice**

**Example text**

He had a great skill in finding examples that served to refute generally believed statements (see pages 309, 426, and 441). In Chapter 1 we often faced the need for a positive integer N that was bigger than some real number. For example, when establishing that lim 1/n = 0 on page 6 we needed N > 1/ε, and we assumed that such an integer exists. Does it? Now we can answer not only that, but even a more general form of the question: if a, b are positive real numbers and a < b, is there a positive integer N such that aN > b?

12. Let A be a non-empty subset of R and let f, g be functions defined on A. (a) Prove that sup{f (x) + g(x) : x ∈ A} ≤ sup{f (x) : x ∈ A} + sup{g(x) : x ∈ A}. (b) Prove that inf{f (x) + g(x) : x ∈ A} ≥ inf{f (x) : x ∈ A} + inf{g(x) : x ∈ A}. (c) Give examples to show that each of the inequalities in (a) and (b) can be strict. 7). Now that we have the Completeness Axiom we can do not only that, but we can also derive several equally important results. We will start with the aforementioned theorem.

6 that the sequence an = (−1)n is divergent. Nevertheless, we have a perfect understanding of its asymptotic behavior (meaning: as n → ∞). When n is an even number, say n = 2k with k ∈ N, then an = 1. On the other hand, when n is an odd number, say n = 2k − 1 with k ∈ N, then an = −1. In order to describe a situation like this we will develop some terminology first. Notice that the sequence a1 , a3 , a5 , . . is convergent. This is not the whole sequence {an }. 1. Let {an } be a sequence, and let n1 < n2 < n3 < .