Combination Theorem for Sequences/Multiple Rule

Theorem

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

Let $\left \langle {x_n} \right \rangle$ be a sequence in $X$ convergent to the following limit:

$\displaystyle\lim_{n \to \infty} x_n = l$

Let $\lambda \in X$.

Then:

$\displaystyle\lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l$

Proof

Let $\epsilon > 0$.

We need to find $N$ such that:

$\forall n > N: \left\vert{\lambda x_n - \lambda l}\right\vert < \epsilon$

If $\lambda = 0$ the result is trivial.

So, assume $\lambda \ne 0$.

Then $\left\vert{\lambda}\right\vert > 0$ from the definition of the modulus of $\lambda$.

Hence $\displaystyle\frac {\epsilon} {\left\vert{\lambda}\right\vert} > 0$.

We have that $x_n \to l$ as $n \to \infty$.

Thus it follows that:

$\exists N: \forall n > N: \left\vert{x_n - l}\right\vert < \dfrac {\epsilon} {\left\vert{\lambda}\right\vert}$

That is:

$\forall n > N: \left\vert{\lambda}\right\vert \left\vert{x_n - l}\right\vert < \epsilon$

But we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{\lambda}\right\vert \left\vert{x_n - l}\right\vert\)	\(=\)	\(\displaystyle \left\vert{\lambda \left({x_n - l}\right)}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Modulus of Product
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert{\lambda x_n - \lambda l}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Hence:

$\displaystyle\lim_{n \to \infty} \left({\lambda x_n}\right) = \lambda l$

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.6 \ (3)$