Let $f:(0, 1)\to\mathbb R$ be such that $f$ is bounded and $\lim\limits_{x\to0} f(x)$ does not exist.

Show that there exists two "null sequences" {$x_n$} and {$y_n$} in $(0, 1)$ such that $\lim\limits_{n\to\infty} f(x_n)$ and $\lim\limits_{n\to\infty} f(y_n)$ both exist but unequal.

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    $\begingroup$ This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. To be more precise: What have you tried so far? Where are you stuck? $\endgroup$
    – AlexR
    Jan 27, 2014 at 11:25
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    $\begingroup$ Find one null sequence $(x_n)$ for which the limit $\lim_{n\rightarrow\infty} f(x_n)$ exists. Call the limit $L$. Find another null sequence $(y_n)$ and an $\epsilon>0$ such that $|f(y_n)-L|>\epsilon$ for all $n$. Now find a convergent subsequence of $(f(y_n))$. Can its limit be $L$? $\endgroup$ Jan 27, 2014 at 11:27
  • $\begingroup$ @user123804 Why what? $\endgroup$
    – AlexR
    Jan 27, 2014 at 11:29
  • $\begingroup$ Not getting David Mitra. Please explain.. $\endgroup$
    – user123804
    Jan 27, 2014 at 11:39
  • $\begingroup$ First recall that any bounded sequence of reals has a convergent subsequence. Using this, you can find $x_n\rightarrow 0$ with $f(x_n)\rightarrow L$ (take any sequence converging to $0$, then find a convergent subsequence of its image under $f$). OK.. Since $\lim_{x\rightarrow0} f(x)\ne L$, you can find a null sequence $(y_n)$ such that $(f(y_n))$ is "bounded away from $L$". That is, there is an $\epsilon>0$ so that for all $n$, $|f(y_n)-L|>\epsilon$. But $(f(y_n))$ has a convergent subsequence. This subsequence can't converge to $L$. $\endgroup$ Jan 27, 2014 at 11:53

1 Answer 1


Hint: Since $f$ is bounded, both $\limsup_{x\to0^+}f(x)$ and $\liminf_{x\to0^+}f(x)$ exist and are finite. (Why?)

Now, if $\limsup_{x\to0^+}f(x)=\liminf_{x\to0^+}f(x)$, what can you say about $\lim_{x\to0^+}f(x)$?

And finally>: if $\limsup_{x\to0^+}f(x)\neq\liminf_{x\to0^+}f(x)$, can you think of a way to leverage this to find the two sequences that you are looking for?


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