The family $f_n=\arctan(nx)$ is not equicontinous. I am studying Equicontinuous families and the book says that $\arctan(nx)$ is not equicontinuous since the definition is violated if $x=0$ I would really appreciate if someone explain me what part of the definition is violated.
My definition is: Let $F$ be a collection of real functions on a metric space $X$ with metric $d$.
We say that $F$ is equicontinuous if to every $\varepsilon>0$ corresponds a $\delta>0$ such that $|f(x)-f(y)|<\varepsilon\ \forall f\in F$ and for all pairs of points $x,y$ with $d(x,y)<\delta$ (In particular, every $f\in F$ is then uniformly continuous.) Thanks in advance!

Denote $f_n(x) = \arctan(nx)$. We know that for $x>0$, $\lim_{n\rightarrow \infty} f_n(x) = \pi/2$; for $x<0$, $\lim_{n\rightarrow \infty} f_n(x) = -\pi/2$. Also, $f_n(0) = 0$. This implies $\exists \epsilon = 1 (<\pi/2)$, $\forall \delta>0$, $\exists N$ and $x\in (-\delta,\delta)$, such that when $n>N$ we have $d(f_n(x),f_n(0)) \ge \epsilon$. This violates the definition of equicontinuity.
