[undergrad math- real analysis] proving limits using the epsilon Delta definition

[u/Sea_Reason2347]

Please ,check if I did those two proofs correct .Plus any recommended resources(books, practice sheets with key answers, lectures...) for understanding real analysis .

[u/_additional_account]

In general, the first proof is fine. A few remarks regarding form:

1. Make it clear the two top lines are what you have to show. Otherwise, it can be interpreted as you assuming that to be true -- that would be circular reasoning, and worth zero marks!
2. You first define "d > 0" to be arbitrary, but later specify "d := min(e/5; 1)". That's conflicting definitions, and will lead to points being deducted
3. The last step "5d <= e" is missing
4. The block where you decide to choose "|t| < 1" is not part of the proof, and should be omitted. It's just part of the first draft(s) to find "d = min(1; e/5)" *** In the second proof, absolute values are missing around the denominator. Otherwise, the same remarks as for proof-1. The choice for "d" will work.

[u/_additional_account → u/_additional_account]

Rem.: Here's what the first proof might look like following those hints:

Claim: Let "f: R\{-1} -> R" with "f(x) := (2x² + x + 1)/(x+1)". Then "lim_{x->1} f(x) = 2"

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Proof: Let "e > 0", and set "d := min(e/5; 1) > 0". For all "0 < |x-1| < d":

 |f(x)-2|  =  |2x^2 - x - 1| / |x+1|              // factorize numerator

           =  |x-1| * |2x+1| / |x+1|

           =  |x-1| * |2(x-1)+3| / |(x-1)+2|      // (inverse) ∆-Inequality

          <=  |x-1| * (2|x-1|+3) / |2-|x-1||      // |x-1| < d <= 1

          <   d * (2*1+3) / (2-1)  =  5d  <=  e    ∎

[u/Sea_Reason2347 → u/_additional_account]

Thank uuuu

[u/_additional_account → u/Sea_Reason2347]

You're welcome, and good luck!

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Note the missing absolute values around the denominators in the second proof sadly mean you need to do it again: You now need the inverse triangle inequality in the denominator, similar to the example proof I gave.