MATA31H3 Lecture Notes - Archimedean Property, Natural Number, Infimum And Supremum
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Please read the following statement and sign below: I understand that any breach of academic integrity is a violation of the code of behaviour on academic matters. By signing below, i pledge to abide by the code. Fix > 0, and suppose x satis es. 0 < |x 2| < min (cid:110) Final exam (1) (a) (15 points) prove that lim x 2 any theorems from lecture. Mata31h3 page 2 of 10 x2 3x + 3 = 1. For this part of the problem you may not use. It follows that |x 2| < 1, i. e. that 1 < x 2 < 1. Thus, for all x satisfying (*), we have 0 < x 1 < 2. On the other hand, (*) implies that |x 2| < . |(x2 3x + 3) 1| = |x 1| |x 2| < 2 .
