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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! :)

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  • $\begingroup$ What if you tried proving the contrapositive instead? $\endgroup$ May 31, 2021 at 11:12
  • $\begingroup$ 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 $f(y_n)$ $\endgroup$
    – Agaman
    May 31, 2021 at 12:02

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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.

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  • $\begingroup$ Thanks! This looks better than mine. :) $\endgroup$
    – Agaman
    May 31, 2021 at 12:04

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