Show that if \((|x_n|) \rightarrow 0\) for all \(n \in \mathbb{N}\), then \(x_n \rightarrow 0\).

Definitions of sequences and convergence: here, Definitions of subsequences: here.
For the “show the limit” template: here.

Alternative Proof (Squeeze Theorem)

To show that \(\lim(x_n) = 0\), let \(\epsilon > 0\) be arbitrary. We want to prove that there exists some \(n \geq N\) such that,

$$ \begin{align*} \lvert x_n - 0 \rvert < \epsilon \\ \lvert x_n \rvert < \epsilon \end{align*} $$

Starting with \(x_n\). We know that \(x_n \leq |x_n|\) for any \(n \in \mathbb{N}\). Similarly \(x_n \geq -|x_n|\) (for the full proof see this. So now we have

$$ \begin{align*} -|x_n| \leq x_n \leq |x_n|. \end{align*} $$

We’re given that \((|x_n|)\) converges to 0. Furthermore, by the algebraic limit theorem we know that if \(\lim(a_n) = a\), then \(\lim(ca_n) = ca\) Therefore \(-|x_n|\) converges to 0 as well by setting \(c\) to -1. so now we have,

$$ \begin{align*} 0 \leq \lim(x_n) \leq 0. \end{align*} $$

By the sequenze theorem for sequences, we can conclude that \((x_n)\) converges to 0 as we wanted to show. \(\blacksquare\)

References