Falsly prove that $\sin(x)$ is discontinuous at $0$?

Question

I've put a lot of time thinking about this question, where one has to prove the discontinuity of

$$ f(x)=\left\{\begin{array}{ll} \sin{\frac{1}{x}}, & x\neq 0 \\ 0, & x= 0\end{array}\right. $$

I know the Satz, that if every subsequence would converge towards 0 it would imply that $\sin{\frac{1}{x}}$ is continuous, but since one can find the subsequence $x_n=\frac{1}{\frac\pi2+n\pi}$ one can refute, that $f(x)$ is continuous going towards $x=0$.

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Now comes the part where I can't seem to find my mistake:

If I take the function $$ g(x)=\left\{\begin{array}{ll} \sin{x}, & x\neq 0 \\ 0, & x= 0\end{array}\right. $$

and define my subsequence as $x_n=\frac{\pi}{2}+n\pi$, couldn't I falsly implicate with that, that $g(x)$ is discontinuous at $x=0$?

I really can't find the point, where my thinking is completely off, so I'd appreciate if someone could clear my misunderstanding.

Thanks in advance.

Answer 1

Note that

$$x_n=\frac{\pi}{2}+n\pi\not \to 0$$

Answer 2

The map $g$ is continuous in $0$ is and only if for each $x_n\to 0$ you have that $g(x_n)\to 0$.

You have chosen a sequence that not converges to zero because

$\frac{\pi}{2}+n\pi\to \infty$

Conversely, in the other case

$\frac{1}{\frac{\pi}{2}+n\pi}\to 0$