Let $$f(x) = \begin{cases}1 \ \ \ \ 1 \le |x| \\ 1/n \ \ \ \ 1/n \le |x| \lt 1/(n-1), \ \ n \in \mathbb{Z} \text{ and } n \ge 2 \\ 0 \ \ \ \ x = 0\end{cases}$$ Prove that $f$ is continuous at $x= 0$ but discontinuous at $x = \pm 1/n$
Let $f^*$ be a hyperreal extension of $f$.
$$\lim_{x \to 0} f^*(x) = st(f^*(|\varepsilon|))$$
Now for any infinitesimal $\varepsilon$ we can find a infinite hyperinteger $K$ such that $${1\over K} \le |\varepsilon| \lt {1\over K-1}$$
Therefore $$ f^*(|\varepsilon|) = {1\over K} = \delta$$ Since $ \lim_{x \to c} f(c) = st(f^*(c))$ $$\therefore \lim_{x \to 0} f(x) = st(f^*(\varepsilon)) = st(\delta) = 0 = f(0)$$
Hence $f$ continous at $0$.
Similarly,
$$\lim_{x \to 1/n} f^*(x) = st(f^*(1/n + \epsilon))$$
Let $n$ be a infinite hyperinterger $H$,
$$f^*(1/H + \varepsilon) = f^*(\delta + \epsilon)$$
Like the proof above we can find a $J$ such that $\displaystyle {1\over J} \le |\delta + \epsilon| \le {1\over J- 1}$
$$\therefore st(f^*(\delta + \epsilon)) = st\left({1\over J}\right) \ne {1\over H} = f^*(1/ H)$$
Therefore $f^*$ is discontinuous at $\pm 1/n$ and therefore $f$ is also discontinuous.
- I getting a feeling like I have used too many symbols that mean nothing and my proof is wrong. I think the second part is probably is wrong. Please tell me if this proof is correct and if not then how to correct it ?
The second part is incorrect. You need to prove discontinuity at $1/n$ for a fixed standard $n$, so it makes no sense to "let $n$ be $H$" as you do here. Here it suffices to point out that for any infinitesimal $\epsilon>0$ you will have $f(\frac{1}{n}-\epsilon)=\frac{1}{n+1}$ which is not infinitely close to $\frac{1}{n}$ and therefore the function is discontinuous at $x=\frac{1}{n}$.