# Combination Theorem for Sequences/Real/Multiple Rule

## Theorem

Let $\sequence {x_n}$ be a sequences in $\R$.

Let $\sequence {x_n}$ be convergent to the following limit:

$\ds \lim_{n \mathop \to \infty} x_n = l$

Let $\lambda \in \R$.

Then:

$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$

## Proof

Let $\epsilon > 0$.

We need to find $N$ such that:

$\forall n > N: \size {\lambda x_n - \lambda l} < \epsilon$

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

So, assume $\lambda \ne 0$.

Then $\size \lambda > 0$ from the definition of the absolute value of $\lambda$.

Hence $\dfrac \epsilon {\size \lambda} > 0$.

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

Thus it follows that:

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

That is:

$\forall n > N: \size \lambda \size {x_n - l} < \epsilon$

But we have:

 $\ds \size \lambda \size {x_n - l}$ $=$ $\ds \size {\lambda \paren {x_n - l} }$ Absolute Value of Product $\ds$ $=$ $\ds \size {\lambda x_n - \lambda l}$

Hence:

$\ds \lim_{n \mathop \to \infty} \paren {\lambda x_n} = \lambda l$

$\blacksquare$