Show that limit exists but are not equal

Question

Problem: Let $f:(0,1)\rightarrow \mathbb{R}$ be bounded but such that $\lim_{x\rightarrow 0} f$ does not exist. Show that there are two sequences $(x_n)$ and $(y_n)$ in $(0,1)$ with $\lim(x_n)=\lim(y_n)=0$ but such that $\lim(f(x_n))$ and $\lim(f(y_n))$ exist but are not equal.

Here's how I approached the problem, I showed the existence of such a sequence $(x_i)$ in $(0,1)$ such that $(f(x_i))$ converges. Can anyone help me with the remaining part or can anyone provide a better solution?

My approach: Let $(x_n)$ be a sequence in $(0,1)$ such that $\lim(x_n)=0$ (such a sequence exists), mow as $f$ is bounded $|f(x_n)|\le M$ for some $M>0$. So by Bolzano-Weierstrass Theorem for sequences, we know that that there is a subsequence $(f(x_i))$ of $(f(x_n))$ such that $(f(x_i))$ is convergent, so the subsequence $(x_i)$ of $(x_n)$ also converges to $0$ as $(x_n)$ converges to $0$, so for the sequence $(x_i)$ in $(0,1)$ the sequence, $\lim(f(x_i))$ exists. Thus,by the sequential criterion, $\lim(f(x_n))$ exists. I don't know how to proceed after this, any help would be really appreciated. Thanks in advance! :)

Answer 1

If $\limsup_{x\rightarrow 0} f(x)$=$\lim\inf_{x\rightarrow 0} f(x)$, then $\lim _{x\rightarrow 0}f(x)$ exists. Since $f$ is bounded, limsup and liminf exist. Since the limit does not exist, they must converge to different things. Now choose sequences converging to each of their limits.