# Show that limit exists but are not equal

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! :)

• What if you tried proving the contrapositive instead? May 31, 2021 at 11:12
• Thanks for your comment @MatthewLeingang , I will definitely try that approach. By the way here's some progress that I got. Does this work? Continuing from where I'd left, suppose $\lim(f(x_i))=L$ as $\lim_{x\rightarrow 0} (y_n)$ doesn't exist we can construct a sequence $(y_n)$ in $(0,1)$ such that there exists some $\epsilon>0$ with the property that $|f(y_n)-L|\ge\epsilon$,so$(f(y_n))$ is bounded,so again applying Bolzano Weierstrass we can get a subsequence,$(f(y_i))$ such that it's convergent,and this definitely doesn't converge to $L$, otherwise it'd contradict the definition of $f(y_n)$ May 31, 2021 at 12:02

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.