Limit laws and inequalities

Try

Suppose a_n approaches 2 and b_n approaches −1. It is natural to expect a_n+b_n to approach 1 and a_nb_n to approach −2. Analysis turns this expectation into reusable error estimates. The sum error separates directly, while the product error contains a varying factor that must be controlled. That difference explains why boundedness appears in the proof rather than being a decorative preliminary lemma.

Predict

Predict whether strict inequalities survive limits. The sequences a_n=0 and b_n=1/n satisfy a_n<b_n at every index, yet both limits are zero. Non-strict order can survive; a fixed positive gap need not. A theorem that concludes L<M from termwise strict inequalities would therefore be false even in this elementary example.

Experiment

Compare the error structures.

Operation	Useful identity or estimate	Extra requirement
Sum	|(a_n+b_n)−(a+b)|≤|a_n−a|+|b_n−b|	Common tail
Product	a_nb_n−ab=a_n(b_n−b)+b(a_n−a)	Bound |a_n|
Reciprocal	|1/b_n−1/b|=|b_n−b|/(|b_n||b|)	b≠0

Use the product decomposition to explain why simply multiplying two small errors would omit terms. Test it with a_n=a+u and b_n=b+v; the full difference is av+bu+uv.

Understand

For the sum law, choose tails with each error below ε/2, then take the maximum cutoff. For the product law, convergent a_n is bounded by some M≥1. Choose |b_n−b|<ε/(2M) and |a_n−a|<ε/(2(|b|+1)). The decomposition bounds the total error by a quantity below ε. The added 1 handles b=0 without dividing by zero.

For reciprocals with b≠0, eventually |b_n−b|<|b|/2, so |b_n|>|b|/2. Hence the reciprocal error is at most 2|b_n−b|/|b|² and tends to zero. There may be zeros in a finite prefix; the reciprocal tail is eventually defined. A quotient law follows by multiplying the numerator sequence by this reciprocal sequence.

Apply

Prove order preservation directly. If a_n≤b_n eventually but a>b, choose ε=(a−b)/3. On a common tail, a_n>a−ε>b+ε>b_n, a contradiction. This proof identifies why a positive limiting gap is impossible.

The squeeze theorem follows: if a_n≤c_n≤b_n eventually and both outer sequences tend to L, then for a prescribed ε the common outer tail lies between L−ε and L+ε. The middle sequence must lie there too. Repair “a sequence bounded between two convergent sequences always converges”: their limits must agree. The alternating sequence lies between the constant sequences −1 and 1 without converging.

Limits also preserve absolute values because ||a_n|−|a||≤|a_n−a|. This reverse triangle inequality supplies a direct error bound, including when the limiting value is zero. Derive it from the ordinary triangle inequality in both directions rather than assuming absolute value may be moved through a limit without justification.

Continue

Optional depth asks which operations preserve limits when viewed as continuous functions of several variables. That perspective will later unify these laws, but using it now without a definition and proof of continuity would be circular.

Save a product proof with the uniform bound and the two tolerance choices visible. Then construct a quotient example whose denominator tends to zero, such as (1/n)/(1/n²)=n, to show why the nonzero-limit hypothesis matters. In the next lesson, completeness will produce limits of monotone sequences and convergent subsequences without an explicit candidate formula.