Theorem (Squeeze Theorem): Let (aₙ), (bₙ), and (cₙ) be sequences such that aₙ → c, bₙ → c and aₙ ≥ cₙ ≥ bₙ for all large enough n. Then cₙ → c.
Proof: Consider an open interval centered at c of radius ε > 0. Since aₙ → c, there must exist some natural number N₁ such that whenever n ≥ N₁, aₙ will be in that open interval. As a consequence, c + ε > aₙ whenever n ≥ N₁.
Something similar happens with (bₙ). Since bₙ → c, there must exist some natural number N₂ such that whenever n ≥ N₂, bₙ will be in that open interval. As a consequence, bₙ > c - ε whenever n ≥ N₂.
By our assumptions, we also know that there is some natural number N₃ such that whenever n ≥ N₃, it follows that aₙ ≥ cₙ ≥ bₙ. Let N be the biggest of these three numbers. Then all three inequalities will be true at the same time, so:
$$c - \varepsilon > a_n \ge c_n \ge b_n > c - \varepsilon$$Which immediately tells us that for our ε > 0, there is a number, namely N, such that whenever n ≥ N, cₙ will be in the open interval centered at c of radius ε. By definition, this means that cₙ → c. ■
Theorem (Const. Multiple Rule): If aₙ → a and c is a constant, then caₙ → ca.
Proof: Let ε > 0. Since aₙ → a, there exists a natural number N₁ such that n ≥ N₁ implies that |aₙ - a| < ε/|c|. It follows that |caₙ - ca| = |c(aₙ - a)| = |c||aₙ - a| < (|c|ε)/|c| = ε. ■
Theorem (Sum Rule of Limits): If aₙ → a and bₙ → b, then aₙ + bₙ → a + b.
Proof: Let ε > 0. Since aₙ → a, there exists a natural number N₁ such that n ≥ N₁ implies that |aₙ - a| < ε/2. Similarly, since bₙ → b, there exists a natural number N₂ such that n ≥ N₂ implies that |bₙ - b| < ε/2. Let N be whichever of N₁ and N₂ is bigger. Then n ≥ N implies that:
$$\begin{align*} |(a_n + b_n) - (a+b)| &= |(a_n - a) + (b_n - b)|\\ &\le |a_n - a| + |b_n - b|\\ &< \varepsilon/2 + \varepsilon/2 = \varepsilon \end{align*}$$Which by definition means that aₙ + bₙ → a + b as desired. ■
Theorem (Product Rule of Limits): If aₙ → a and bₙ → b, then aₙbₙ → ab.
Proof: First, let's prove the special case where (aₙ) and (bₙ) are both evanescent. Since convergent sequences are bounded, (aₙ) is bounded. Since the product of a bounded sequence and an evanescent sequence is evanescent, the sequence (aₙbₙ) is evanescent.
To prove the more general case, it suffices to show that if aₙ - a → 0 and bₙ - b → 0, then aₙbₙ - ab → 0. We can achieve this by strategically adding zero a couple times and applying limit properties we have already proven. See below:
$$\begin{align*} a_nb_n-ab &= a_nb_n-a_nb+a_nb-ab\\ &= a_n(b_n-b)+(a_n-a)b\\ &= (a_n-a+a)(b_n-b)+b(a_n-a)\\ &= (a_n-a)(b_n-b)+a(b_n-b)+b(a_n-a) \end{align*}$$Since aₙ - a → 0 and bₙ - b → 0, the fact that (aₙ - a)(bₙ - b) → 0 follows from the special case we just proved. By the constant multiple rule of limits, we also know that a(bₙ - b) → 0 and b(aₙ - a) → 0. Finally, by the sum rule of limits, the whole final line must converge to zero, so aₙbₙ - ab → 0. ■
Theorem (Quotient Rule of Limits): If aₙ → a and bₙ → b ≠ 0, then aₙ/bₙ → a/b.
Proof: We will first prove the special case where aₙ = 1. Since b is not zero, it is possible to put an open interval centered at b where one of the endpoints is exactly between b and zero. This open interval will have a radius of |b|/2.
Because bₙ → b, we know that there exists a natural number N such that when n ≥ N, all of the bₙ will fall inside this interval. Consequently, the distance from 0 to bₙ will be greater than |b/2|, or symbolically, |bₙ| > |b/2|. Flipping the fractions shows that |1/bₙ| < |2/b|. Thus, the following string of inequalities must hold if n ≥ N.
$$0 < \left|\frac{1}{b_n} - \frac{1}{b}\right| = \left|\frac{b_n-b}{b_nb}\right| = \left|\frac{1}{b}\right|\left|\frac{1}{b_n}\right|\left|b_n-b\right| < \frac{2}{b^2}|b_n-b|$$We already know that |bₙ - b| → 0, so by the constant multiple rule, since 2/b² is a constant, 2|bₙ - b|/b² → 0. Thus, by the squeeze theorem, |(1/bₙ) - (1/b)| → 0, which is the same thing as 1/bₙ → 1/b. The general case then follows from the product rule. ■
Theorem: Let aₙ → a and bₙ → b be sequences such that aₙ > bₙ for all large enough n. Then a ≥ b.
Proof: Let aₙ → a and suppose that a < 0. We can put an open interval centered at a that goes all the way to 0 by making its radius |a|. Since aₙ → a, there must be a natural number N such that when n ≥ N, aₙ is in the interval. However, since that interval is bounded above by 0, when n ≥ N it follows that aₙ ≤ 0.
This means that if aₙ → a and aₙ > 0 for every n, then a ≥ 0. After all, we just proved that if a < 0, then it couldn't be true that aₙ > 0 for every n in the first place. Thus, we have proven a special case of the theorem. (DONT FORGET TO FINISH PROOF USING TAILEND THEOREM)
Let's move on to a more general case. Suppose that aₙ > bₙ for all n. Then aₙ - bₙ > 0, so by the special case we just proved, as well as the limit properties we proved earlier, it follows that a - b > 0, and thus that a > b. ■