# Sequence convergence and limits

if I have a sequence $(x_n)_{n\geq0}$ such that $\lim_{n \to\infty } x_n = c$. Then why can I write: $\left | c-x_n \right | = \lim_{m \to\infty } \left | x_m - x_n \right |$

I tried to work it out but I get an inequality: $\left | c-x_n \right | \leqslant \left | c-x_m \right | + \left | x_m-x_n \right |$ Then as m tends to infinity the first term of the right side tends to zero. Is this the right way? What am I missing that makes both sides equal?

• why the down vote? Jan 6, 2016 at 19:39

Because the function $f(u) = |u - x_n|$ is continuous.

Remember that $f$ is continuous in a point $c$ if and only if for every sequence $(x_m)_{m \ge 1}$ with $x_m \to c$ we have that $\lim \limits _{m \to \infty} f(x_m) = f(\lim \limits _{m \to \infty} x_m)$ (the last number being, of course, $f(c)$). Now, apply this to the $f$ given on the first line.

Note that this description of continuity is not the definition (which would be uncomfortable to use here), but a theorem.

• Sorry I don't get it Jan 6, 2016 at 19:39
• Have you studied continuity? If not, it's normal not to understand my answer. Jan 6, 2016 at 19:41
• yes I have. What I understand from your answer is that the limit of $|u - x_n|$ as u tends to c is equal to $f(c)$ Jan 6, 2016 at 19:45

Since $x_m \to c$, we have for any $\epsilon > 0$, $|x_m -c| < \epsilon$ when $m$ is sufficiently large.

Using the reverse triangle inequality,

$$||x_m-x_n|-|c-x_n|| = ||x_m-x_n|-|x_n-c|| \\\leqslant |(x_m-x_n)+(x_n-c)| \\= |x_m-c| < \epsilon.$$

Hence,

$$\lim_{m \to \infty}|x_m - x_n| = |c - x_n|$$