Demonstration analysis and good books?

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SUMMARY

The discussion focuses on the limit calculation lim (x^2) as x approaches 2, specifically addressing the delta-epsilon definition of limits. The user encountered a discrepancy in delta values between their calculations and those presented in Demidovich's book, which suggests delta = epsilon/5. Additionally, the user seeks recommendations for beginner-friendly books on algebra, calculus, and real analysis, particularly those that cover polar coordinates. The consensus emphasizes the necessity of completing the calculus sequence before undertaking an analysis course.

PREREQUISITES
  • Understanding of limit definitions in calculus
  • Familiarity with delta-epsilon proofs
  • Basic knowledge of algebra and functions
  • Concepts of polar coordinates in mathematics
NEXT STEPS
  • Research the delta-epsilon definition of limits in calculus
  • Study the book "Demidovich's Problems in Mathematical Analysis" for limit problems
  • Explore beginner textbooks on real analysis, such as "Understanding Analysis" by Stephen Abbott
  • Learn about polar coordinates and their applications in calculus
USEFUL FOR

Students in mathematics, particularly those studying calculus and real analysis, as well as educators seeking effective teaching resources for these subjects.

fgyamauti
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When I try to demonstrate that lim (x^2)=4
x--->2

I found a different delta (delta=min{2-sqrt(epsilon-4),sqrt(epsilon+4)-2}, towards the one that is written in Demidovich´s book (delta=epsilon/5). Could someone help me?
Could someone tell me, too, a good algebra, calculus and real analysis book? Is it ok if i try analysis next semester absent a more advanced calculus (without calculus 2)? I need some books that explain polar coordinates too (good ones for beginner).
Thanks and sorry for my mathematical notation.
 
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You want [itex]|x^2- 4|= |(x- 2)(x+ 2)|= |x+ 2||x- 2|< \epsilon[/itex].
So we can write [itex]|x- 2|< \epsilon/|x+ 2|[/itex]
But we need a value, [itex]\delta[/itex] that does NOT depend on x so that is [itex]|x- 2|< \delta[/itex] then [itex]|x- 2|< \epsilon/|x+ 2|[/itex]. That means we want [itex]\delta< \epsilon/|x+2|[/itex].

If |x- 2|< 1 (The "1" is chosen just because it is simple. Any positive number would do.) then -1< x- 2< 1 so 1< x< 3 and then 3< x+2< 5 so that 3< |x+ 2|< 5. Then
[tex]\frac{1}{5}< \frac{1}{|x+2|}< 1/3[/tex]
and so
[tex]\frac{\epsilon}{5}< \frac{\epsilon}{|x+2|}[/tex]

That is, if
[tex]|x- 2|< \frac{\epsilon}{5}[/tex]
and |x- 2|< 1 as assumed above to get this, we will have
[tex]|x-2|< \frac{\epsilon}{|x+2|}[/tex]
so
[tex]|x-2||x+2|= |x^2- 4|< \epsilon[/tex]
as need.

In other words, take [itex]\delta[/itex] to be the smaller of [itex]\epsilon/5[/itex] or 1.

Analysis is basically the theory behind the Calculus. I would NOT take an analysis course without having completed the Calculus sequence.
 

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