Bounded function on $(0,1)$ but discontinuous at $0$ Problem:
Let$\ $ $f:(0,1)\rightarrow \mathbb{R}$$\ $ be bounded but such that $\lim_{x\rightarrow 0}f(x)$ does not exist. I need to show that there are two sequences ($x_{n}$) and ($y_{n}$) such that $\lim_{n\rightarrow \infty }x_{n}=\lim_{n\rightarrow \infty }y_{n}=0$ but $\lim_{n\rightarrow \infty }f(x_{n})$ and  $\lim_{n\rightarrow \infty }f(y_{n})$ exist and are not equal.
Here is what I did: We know by density theorem that we can construct a sequence of rational numbers ($x_{n}$) that converges to $0$. Similarly, I can construct another sequence of irrational numbers that converges to $0$. I tried to use the boundedness of $f$ and the Bolzano Weierstrass theorem, but I couldn't reach any conclusion. Any help is appreciated!
 A: Using rational and irrational number is not going to be fruitful, since there's no relation known between the values of $f$ on rationals and values on irrationals. (As is often the case, the function $f$ which is $1$ on the rationals and $0$ on the irrationals shows that this approach is problematic).
For a different approach: We know that the limit does not exist. Choose any sequence $x_n \to 0$. If $f(x_n)$ does not converge, we're done (choose subsequences appropriately). Otherwise, $f(x_n)$ converges to $L$.
However, can you use the fact that $f$ does not have a limit at zero to find points arbitrarily close to zero that are not close to $L$? Can you use this idea to finish the proof?
A: Let $g$ be any bounded function such that $g(0) \ne 0$. Let $f(x) = 0 $ if $x$ is rational and $f(x) = g(x)$ if $x$ is irrational. Won't your method work?
Also why do you need boundedness? All you really need is how $f$ behaves close to $x=0$? I must be missing something.
A: Consider
$$f(x)=\sin(\frac{1}{x}).$$
It is clearly bounded in $(0,1)$.
Take $x_n=\frac{1}{2n\pi}$ so $f(x_n)=0$. Take $y_n=\frac{1}{2n\pi+\frac{\pi}{2}}$, we have $f(y_n)=1$.
As for the proof a counter argument should do the job just fine:
If for any two sequences $x_n\to0$, $y_n\to0$, $\lim_{n→\infty} f(x_n)=\lim_{n→\infty} f(y_n)=K$, That means $\lim_{x→0}f(x)=K$.
