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Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function and $A \subset \mathbb R$ be defined by $A=\{y \in \mathbb R:y=\lim _{n\rightarrow \infty}f(x_n),$for some sequence $x_n\rightarrow \infty\}$

Then the set $A$ is necessarily

A.a connected set

B.compact set

C. a singleton set

D.None of above

Now since $f$ is a continuous function and $x_n$ diverges so $f(x_n)$ will also diverge.Hence the set can't be bounded and hence not compact

Also it may not be singleton as $f(x_n)$ may be either $\infty$ or -$\infty$

Not sure with A.Can someone please check my solution and suggest required edits

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Continuous functions take convergent sequences to convergent sequences, but you can't say the same for divergent sequences. For instance, consider a constant function. For a better example, which will also give a hint for how to solve your problem, take $f(x) = \sin(x)$. For a still better example, think of how you could modify this $f$ to change the set $A$...

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  • $\begingroup$ May be $x_n=n\pi$ divergent but $f(x_n)$ is not.But I did not get what you are asking about modification $\endgroup$ – Learnmore Jan 15 '15 at 14:47
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    $\begingroup$ @learnmore: if $f(x)=\sin(x)$, then $A=[-1,1]$, which is both compact and connected. But if you modify $f$ slightly, you can make $A$ non-compact. Think about that for an hour or two before coming straight back with "I don't get it". $\endgroup$ – TonyK Jan 15 '15 at 14:56
  • $\begingroup$ can u please give some more hints about the modification @TonyK $\endgroup$ – Learnmore Jan 15 '15 at 15:22
  • $\begingroup$ @learnmore: I said an hour or two, not half an hour. Put some effort in. $\endgroup$ – TonyK Jan 15 '15 at 16:16
  • $\begingroup$ should I modify $f$ such that its image becomes $[-1,1)$ or $(-1,1]$ which is not closed and hence not compact @TonyK $\endgroup$ – Learnmore Jan 16 '15 at 2:34

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