Limit point of a given set: proof verification

Question

Let $A = \{(x,y): y = \sin(\frac{1}{x}), 0 < x \le 1\}$ Show that the point $(0, \frac {1}{\sqrt{2}})$ is a limit point of $A$

If $p$ is a limit point of $A$, then $\forall_{\delta>0}, B_\delta(p) \cap A \neq \emptyset$, so we need to show that definition with $p = (0, \frac {1}{\sqrt{2}})$.

My question is how do I proceed to prove this proposition.

Answer 1

For any $p \in [-1,1]$ we have that for some $t \in [0,2\pi]$ we have that $\sin(t) = p$, and so also $\sin(t+2n\pi) = p$ for all $n \ge 1$ (periodicity).

Hence $(\frac{1}{t+2n\pi}, p) \in A$ for all $n \ge 1$ and as $n \to \infty$ we see that this sequence of points of $A$ tends to $(0,p)$, so $(0,p)$ is a limit point of $A$ for all $p \in [-1,1]$.

Answer 2

Hint: $\left(\frac4\pi,\frac1{\sqrt2}\right)\in A$. And $\left(\frac4{5\pi},\frac1{\sqrt2}\right)\in A$ too.