Then dominate it by something simple: for n≥2, 2n2−3≥n2, so the error is ≤2n25≤2n5. Given ε>0, the Archimedean property provides N≥max(2,2ε5); for n≥N the error is ≤ε. Done. The closing insight: an ε–N proof has exactly three moves — compute the error, bound it by a decreasing elementary expression, solve for the threshold — and after this chapter’s theorems (operations, squeeze) one almost never writes such a proof again: the theorems package the three moves once and for all.
Example 11.3(Divergence to infinity, certified)
Claim: un=n2−100n→+∞. Factor the dominant term: un=n2(1−n100)≥2n2 for n≥200. Given M, take N=max(200,⌈2M⌉): for n≥N, un≥2n2≥M. Two habits are on display: dominant-term factoring converts a competition (n2 against −100n) into a single scale times a factor tending to 1; and the threshold may be huge (u100=0, the sequence is even negative before n=100) — divergence to +∞ is a statement about the tail, indifferent to any finite amount of misbehavior.
Example 11.6(Operations plus one algebraic trick)
Compute lim(n2+n−n). The two pieces separately tend to +∞: the operations theorem says nothing about their difference (an indeterminate form). Multiply by the conjugate:
n2+n−n=n2+n+n(n2+n)−n2=n2+n+nn=1+n1+11.
Now everything converges: 1+n1→1, because 0≤1+h−1=1+h+1h≤h (conjugate again, then squeeze with h=n1); then the operations theorem gives the limit 1+11=21. The closing insight: the operations theorem is not a calculator for all limits — indeterminate forms (∞−∞, 00, 0×∞, 1∞) must first be transformed by algebra (conjugates, factoring the dominant term) until every piece converges; the systematic machine for resistant cases is the asymptotic expansion of Chapter 16.