The Algebraic Limit Theorem
Let \(\lim x_n = x\). Then, \(\lim (\alpha x_n) = \alpha x\) for all \(\alpha \in \mathbb{R}\).

Strategy/Notes

For (i) we want to show that \(\lim (\alpha x_n) = \alpha x\). So for any \(\epsilon > 0\), we want to find \(N \in \mathbb{N}\) such that when \(n \geq N\), then

$$ \begin{align*} \big\lvert \alpha x_n - \alpha x \big\rvert < \epsilon \end{align*} $$

Solving for \(n\):

$$ \begin{align*} \big\lvert \alpha x_n - \alpha x \big\rvert &< \epsilon \\ \big\lvert \alpha \big\rvert \big\lvert x_n - x \big\rvert &< \epsilon \\ \big\lvert x_n - x \big\rvert &< \frac{\epsilon}{|\alpha|}. \end{align*} $$

So now we are ready to write a formal proof.

Proof

Let \(\epsilon > 0\) be given. If \(\alpha = 0\), then the result follows trivially since \(\alpha x_n = \alpha x = 0\) for all \(n \in \mathbb{N}\). So we may assume \(\alpha \neq 0\). Now, since \(\lim x_n = x\), then by the definition of limit, there exists a natural number \(N \in \mathbb{N}\) such that for all \(n \geq N\), we must have

$$ \begin{align*} |x_n - x| &< \frac{\epsilon}{|\alpha|}. \end{align*} $$

We now verify that this choice has the desired property. Let \(n \geq N\). Then,

$$ \begin{align*} \big\lvert \alpha x_n - \alpha x \big\rvert = \big\lvert \alpha \big\rvert \big\lvert x_n - x \big\rvert < \big\lvert \alpha \big\rvert \frac{\epsilon}{|\alpha|} = \epsilon. \\ \end{align*} $$

Thus,

$$ \begin{align*} \big\lvert \alpha x_n - \alpha x \big\rvert < \epsilon. \\ \end{align*} $$

as required. \(\blacksquare\)

References