The absolute value is compatible with limits

Let $(x_k)$ be a sequence of reals converging to $x \in \mathbb{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). Then $(|x_k|)$ converges to $|x|$.

In the single case $x = 0$ the implication reverses: $|x_k| \to 0$ if and only if $x_k \to 0$. Whether the implication can be reversed for $x \ne 0$ is taken up in the remarks below; it is no part of what the proof establishes.

• The converse fails at every nonzero limit. This is not established by the proof above, which proves only the forward implication and the equivalence at $x = 0$; the witness is exhibited here instead. Fix a real $c \ne 0$, let $(s_k)$ be the alternating sequence of $1$ and $-1$ constructed in FALSE: every bounded sequence converges, which is shown there not to converge, and put $w_k := s_k c$. Then $|w_k| = |s_k|\,|c| = |c|$ for every $k$ (Basic properties of the absolute value), so $(|w_k|)$ is the constant sequence $|c|$ and converges to $|c|$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences). But $(w_k)$ does not converge: if it converged to some $\ell$, then $(s_k) = (c^{-1}w_k)$ would converge to $c^{-1}\ell$ by the scalar-multiple rule (Algebra of limits: sums, scalar multiples, products and quotients), which it does not. Passing to absolute values destroys sign information, and only at $0$ is there no sign information to destroy.

• Combined with Algebra of limits: sums, scalar multiples, products and quotients this gives the usual companions: the identities $\max\{u, v\} = \tfrac{1}{2}\bigl(u + v + |u - v|\bigr)$ and $\min\{u, v\} = \tfrac{1}{2}\bigl(u + v - |u - v|\bigr)$ (Maximum and minimum of a set), each a two-case check on the sign of $u - v$, exhibit $\max\{x_k, y_k\}$ and $\min\{x_k, y_k\}$ as sums of convergent sequences, so they converge to $\max\{x, y\}$ and $\min\{x, y\}$.

• The lemma is the sequential form of the statement that $u \mapsto |u|$ is continuous, but continuity is not available yet and is not needed: the reverse triangle inequality does the work directly.