Show that $\{\sin(nx) \; \mid \; n \in N\}$ is not equicontinuous at $x=1$

Question

Show that $\{\sin(nx) \; \mid \; n \in N\}$ is not equicontinuous at $x=1$

I have not fully grasped the concept of equicontinuity and uniform equicontinuity. If someone can solve this properly it will really help me get more comfortable with it

$H\subset C(X)$ is equicontinuous at $ x_0$ if $\forall \epsilon>0 \; \exists \delta(x,\epsilon)\in N \ni d(x,x_0)<\delta \rightarrow |f(x)- f(x_0)|< \epsilon \; \forall f\in C(X)$

Answer 1

Suppose that $\{ \sin {nx} \}$ is equicontinuous.
Then, for $ \epsilon = \frac{1}{2}$, $\exists \delta >0 $ such that if $d( x, x_0 ) < \delta$, we have $\sin nx - \sin n < \frac{1}{2}$.

How do we get a contradiction? If you get stuck, look at the following hints.

Take $n$ large enough so that $ n \delta >> 2\pi$.

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Show that there exists $x_1, x_2 \in (x_0 - \delta, x_0 + \delta)$ such that $ \sin nx_1 = 1, \sin nx_2 = -1$.

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Why does this lead to a contradiction?