Limit of Functions proof; help

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SUMMARY

The discussion centers on proving that for a bounded real-valued function \( f \) on a set \( E \subset \mathbb{R} \) with a limit point \( p \), if \( \lim_{x \to p} f(x) \) does not exist, then there exist sequences \( p_n \) and \( q_n \) in \( E \) converging to \( p \) such that \( \lim_{n} f(p_n) \) and \( \lim_{n} f(q_n) \) exist but are distinct. Participants emphasized the necessity of constructing these sequences carefully to avoid convergence issues, particularly when \( E \) contains specific forms like \( 1/n \).

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Homework Statement



Let E [tex]\subset \mathbb{R}[/tex] and E [tex]\neq \emptyset[/tex]. Fix p as a limit point of E. Suppose that f is bounded and realvalued on E and that [tex]\lim_{x \to p}f(x)[/tex] does not exist. Prove the fact that there exist sequences p_n and q_n in E with [tex]\lim_{n}p_n=\lim_{n}q_n=p[/tex] such that [tex]\lim_{n}f(p_n)[/tex] and [tex]\lim_{n}f(q_n)[/tex] exist, but are different.


Homework Equations





The Attempt at a Solution



Since p is a limit point of E, we are guaranteed a sequence (p_n) in E where [tex]p_n \neq p[/tex] and [tex]\lim_{n}p_n=p[/tex]. Using the same logic I can find another sequence, call it (q_n), that converges to p.

Now I am unsure how [tex]\lim_{n}f(p_n)[/tex]and [tex]\lim_{n}f(q_n)[/tex] exist?
 
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f is bounded, so given a sequence of values (f(pn), you can find a convergent subsequence. Use the subsequence instead of the full sequences you originally find.

I'd be wary about the 'using the same logic' part though... let's say E is the set of all numbers of the form 1/n... if my first sequence is an = 1/n, I've screwed myself out of finding any more sequences. You need to work your way around this problem
 

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