Limit Laws (Algebraic Limit Theorem)

Algebraic Limit Theorem

Let $(a_{n}), (b_{n})$ be convergent sequences, where $a_{n} \to a, b_{n} \to b$ . Then:

For $c \in R$ , $lim (c a_{n}) = c a$ .
$lim (a_{n} + b_{n}) = a + b$
$lim (a_{n} b_{n}) = a b$
$lim (\frac{a_{n}}{b_{n}}) = \frac{a}{b}$ given $b \neq 0$ . (ignore the cases when $b_{n} = 0$ for the inner sequence. At most finitely many $b_{n} = 0$ )

Scratch Work:

We want, for some $ε > 0$ :

| c a_{n} - c a | < ε \Rightarrow | c | | a_{n} - a | < ε

Consider the cases. If $c = 0$ then the case is trivial. If $c \neq 0$ then:

| a_{n} - a | < \frac{ε}{| c |}

We know that $\exists N$ such that for all $n \geq N$ .

We want, for some $ε > 0$ :

| (a_{n} + b_{n}) - (a + b) | < ε

Notice then:

\begin{aligned} | (a_{n} - a) + (b_{n} - b) | & \leq \underset{< \frac{ε}{2}}{\underset{⏟}{| a_{n} - a |}} + \underset{< \frac{ε}{2}}{\underset{⏟}{| b_{n} - b |}} \end{aligned}

We want, for some $ε > 0$ :

| a_{n} b_{n} - a b | < ε

Notice that, ignoring the absolute values for a moment:

\begin{aligned} a_{n} b_{n} - a b - a b_{n} + a b_{n} & = (a_{n} - a) b_{n} + a (b_{n} - b) \end{aligned}

So putting the absolute values back in:

\begin{aligned} | (a_{n} b_{n}) - a b | & = | (a_{n} - a) b_{n} + a (b_{n} - b) | \\ \leq | b_{n} | \underset{< \frac{ε}{2 M}}{\underset{⏟}{| a_{n} - a |}} + | a | \underset{< \frac{ε}{2 (| a |)} \lor \frac{ε}{2 (| a | + 1)}}{\underset{⏟}{| b_{n} - b |}} & Triangle Ineq. \end{aligned}

where since $b_{n}$ is convergent, then it has a bound $M$ , so then $| b_{n} | \leq M$ . Notice further that $M > 0$ since we consider the absolute value of $b_{n}$ .

Say $b \neq 0$ . Then we want:

\frac{1}{b_{n}} \to \frac{1}{b}

Because then we can apply (3) by using the new sequence $1 / b_{n}$ instead. This means we need to show:

| \frac{1}{b_{n}} - \frac{1}{b} | = \frac{| b - b_{n} |}{| b | | b_{n} |}

Note that because $(b_{n}) \to b$ , then we can make the numerator as small we we like by making $n$ large. On the other hand, for the denominator we want some $| b_{n} | \geq δ > 0$ . This will lead to a bound on the size of $\frac{1}{| b | | b_{n} |}$ .

The trick is to look far enough out into the sequence $(b_{n})$ so that the terms are closer to $b$ than they are to $0$ . Consider $ε_{0} = \frac{| b |}{2}$ . Because $(b_{n}) \to b$ then there exists and $N_{1}$ such that $| b_{n} - b | < \frac{| b |}{2}$ for all $n \geq N_{1}$ . This implies that $| b_{n} | > \frac{| b |}{2}$ . So then choose $N_{2}$ such that $N \geq N_{2}$ implies:

| b_{n} - b | < \frac{ε | b |^{2}}{2}

and choose $N = max (N_{1}, N_{2})$ .

Proof

Let $ε > 0$ . Choose $N \in N$ such that:
a. If $c = 0$ then we are done (trivial) case.
b. Since $c \neq 0$ then because $\frac{ε}{| c |} > 0$ then $n \geq N \Rightarrow | a_{n} - a | < \frac{ε}{| c |}$ . Then:

\begin{aligned} | a_{n} - a | & < \frac{ε}{| c |} \\ | c | | a_{n} - a | & < ε \\ | c a_{n} - c a | & < ε \end{aligned}

Let $ε > 0$ . Since $\frac{ε}{2} > 0$ then $\exists N_{1}$ where $n \geq N_{1}$ implies $| a_{n} - a | < \frac{ε}{2}$ . Similarly then $\exists N_{2}$ where $n \geq N_{2}$ implies $| b_{n} - b | < \frac{ε}{2}$ . Choose $N = max (N_{1}, N_{2}) \in N$ , since then letting $n \geq N$ :

\begin{aligned} | (a_{n} + b_{n}) - (a + b) & = | (a_{n} - a) + (b_{n} - b) | \\ \leq | a_{n} - a | + | b_{n} - b | & Δ Inequality \\ < \frac{ε}{2} + \frac{ε}{2} \\ = ε \end{aligned}

Let $ε > 0$ . Since $b_{n} \to b$ then $\exists M \geq 0$ where $| b_{n} | \leq M$ . Notice that:

ε > 0 \Rightarrow \frac{ε}{2 M} > 0

And further:

ε > 0 \Rightarrow \frac{ε}{2 (| a | + 1)} > 0

(or choose $ε / 2 (| a |)$ if $| a | \neq 0$ . Either way choosing either doesn't affect the later arguments). Then since $a_{n} \to a$ then $\exists N_{1}$ where for all $n \geq N_{1}$ then $| a_{n} - a | < \frac{ε}{2 (| a | + 1)}$ . Similarly since $b_{n} \to b$ then $\exists N_{2}$ where for all $n \geq N_{2}$ when $| b_{n} - b | < \frac{ε}{2 M}$ .

Now choose $N = max (N_{1}, N_{2})$ . Let $n \geq N$ . Then:

\begin{aligned} | a_{n} b_{n} - a b | & = | a_{n} b_{n} - a b_{n} + a b_{n} - a b | \\ = | (a_{n} - a) b_{n} + a (b_{n} - b) | \\ \leq | b_{n} | | a_{n} - a | + | a | | b_{n} - b | \\ < \frac{| b_{n} | ε}{2 M} + \frac{| a | ε}{2 (| a | + 1)} \\ < \underset{| b_{n} | \leq M}{\underset{⏟}{\frac{ε}{2}}} + \underset{| a | < \frac{ε}{2 (| a | + 1)}}{\underset{⏟}{\frac{ε}{2}}} \\ = ε \end{aligned}

If we can prove that $\frac{1}{b_{n}} \to \frac{1}{b}$ then we can apply (3) and we are done (assuming $b \neq 0$ ). Let $ε > 0$ . First note:

| \frac{1}{b_{n}} - \frac{1}{b} | = \frac{| b - b_{n} |}{| b | | b_{n} |}

Now because $(b_{n}) \to b$ then $| b_{n} - b | < \frac{ε | b |^{2}}{2}$ since it's $> 0$ . There's some $N_{2}$ such that $n \geq N_{2}$ allows for this inequality to hold.

Now notice also that because $(b_{n}) \to b$ then notice for a specific $ε_{0} = \frac{| b |}{2}$ that then $\exists N_{1}$ where $| b_{n} - b | < \frac{| b |}{2}$ for all $n \geq N_{1}$ . Then this implies that $| b_{n} | > \frac{| b |}{2}$ .

Thus choose $N = max (N_{1}, N_{2})$ so then for all $n \geq N$ :

| \frac{1}{b_{n}} - \frac{1}{b} | = | b - b_{n} | \frac{1}{| b | | b_{n} |} < \frac{ε | b |^{2}}{2} \cdot \frac{1}{| b | \frac{| b |}{2}} = ε

☐