Algebraic_Limit_Theorem

Let $\lim a_{n} = a$ and $\lim b_{n} = b$. Then,

  1. $\lim\left. \left( ca_{n} \right) \right. = ca$, for all $c \in \mathbb{R}$

  2. $\lim\left. \left( a_{n} + b_{n} \right) \right. = a + b$

  3. $\lim\left. \left( a_{n}b_{n} \right) \right. = ab$

  4. $\lim\left. \left( a_{n}/b_{n} \right) \right. = a/b$, provided $b \neq 0$

Proof of (1)

Consider the case where $c \neq 0$. We want to show that the sequence $\left. \left( ca_{n} \right) \right.$ converges to $ca$. Let $\epsilon > 0$ be arbitrary. Our goal is to find some point in the sequence $\left. \left( ca_{n} \right) \right.$ after which we have

$$ \left| ca_{n} - ca \right| < \epsilon $$

Now,

$$ \left| ca_{n} - ca \right| = |c|\left| a_{n} - a \right| $$

We are given that $\left. \left( a_{n} \right) \right. \rightarrow a$, so we know we can make $\left| a_{n} - a \right|$ as small as we like. In particular, we can choose an $N$ such that

$$ \left| a_{n} - a \right| < \frac{\epsilon}{|c|} $$

whenever $n \geq N$. To see that this $N$ indeed works, observe that, for all $n \geq N$,

$$ \left| ca_{n} - ca \right| = |c|\left| a_{n} - a \right| < |c|\frac{\epsilon}{|c|} = \epsilon $$

The case $c = 0$ reduces to showing that the constant sequence $\left. (0,0,0,\ldots) \right.$ converges to 0, which is easily verified.

Proof of (2)

To prove this statement, we need to argue that the identity

$$ \left| \left. \left( a_{n} + b_{n} \right) \right. - \left. (a + b) \right. \right| $$

can be made less than an arbitrary $\epsilon$ using the assumptions that $\left| a_{n} - a \right|$ and $\left| b_{n} - b \right|$ can be made as small as we like for large $n$. The first step is to use the triangle inequality to say

$$ \left| \left. \left( a_{n} + b_{n} \right) \right. - \left. (a + b) \right. \right| = \left| \left. \left( a_{n} - a \right) \right. + \left. \left( b_{n} - b \right) \right. \right| \leq \left| a_{n} - a \right| + \left| b_{n} - b \right| $$

Again, we let $\epsilon > 0$ be arbitrary. The technique this time is to divide the $\epsilon$ between the two expressions on the right-hand side in the preceding inequality. Using the hypothesis that $\left. \left( a_{n} \right) \right. \rightarrow a$, we know there exists an $N_{1}$ such that

$$ \left| a_{n} - a \right| < \frac{\epsilon}{2}\quad\mathrm{\text{whenever }}n \geq N_{1} $$

Likewise, the assumption that $\left. \left( b_{n} \right) \right. \rightarrow b$ means that we can choose an $N_{2}$ such that

$$ \left| b_{n} - b \right| < \frac{\epsilon}{2}\quad\mathrm{\text{whenever }}n \geq N_{b} $$

The question now arises as to which of $N_{1}$ or $N_{2}$ we should take to be our choice of $N$. By choosing $N = \max\{ N_{1},N_{2}\}$, we ensure that if $n \geq N$, then $n \geq N_{1}$ and $n \geq N_{2}$. This allows us to conclude that

$$ \begin{aligned} |\left. \left( a_{n} + b_{n} \right) \right. - \left. (a + b) \right. & \leq \left| a_{n} - a \right| + \left| b_{n} - b \right| \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ & = \epsilon \end{aligned} $$

for all $n \geq N$, as desired.