Combination Theorem for Sequences/Real/Product Rule

Theorem
Let $X$ be one of the standard number fields $\Q, \R, \C$.

Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be sequences in $X$.

Let $\left \langle {x_n} \right \rangle$ and $\left \langle {y_n} \right \rangle$ be convergent to the following limits:


 * $\displaystyle \lim_{n \mathop \to \infty} x_n = l, \lim_{n \to \infty} y_n = m$

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} \left({x_n y_n}\right) = l m$

Proof
Because $\left \langle {x_n} \right \rangle$ converges, it is bounded by Convergent Sequence is Bounded.

Suppose $\left\vert{x_n}\right\vert \le K$ for $n = 1, 2, 3, \ldots$.

Then:

But $x_n \to l$ as $n \to \infty$.

So $\left\vert{x_n - l}\right\vert \to 0$ as $n \to \infty$ from Convergent Sequence Minus Limit.

Similarly $\left\vert{y_n - m}\right\vert \to 0$ as $n \to \infty$.

From the Combined Sum Rule:
 * $\displaystyle \lim_{n \mathop \to \infty} \left({\lambda x_n + \mu y_n}\right) = \lambda l + \mu m$, $z_n \to 0$ as $n \to \infty$

The result follows by the Squeeze Theorem for Sequences of Complex Numbers (which applies as well to real as to complex sequences).