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**Isomorphism** That solution you got is of no use to part A. To demonstrate discontinuity at $\displaystyle x_0$, we have to find a sequence $\displaystyle x_n \to x_0$, but the sequence $\displaystyle f(x_n) \not \to f(x_0)$.

Your sequence $\displaystyle x_n \to 0$, but $\displaystyle f(x_n) = \sin 2n \pi = 0$. So it is of no use.

As a side note, I am really sorry transgalactic, that I am failing in explaining this concept to you. I have already given you the answer for this in my previous posts. The sequence you **can** use is $\displaystyle x_n = \frac2{(2n+1)\pi}$

I am telling the following based on our discussion.....If you take it in the right spirit, I suggest you take tutions. Sometimes a live tutor teaching a new concept is an invaluable experience.