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I need help in order to confirm whether my proof is approved or not. It follows as:

Claim: Let $f: [a,\infty) \mapsto \mathbb{R}$ where $f$ is continous. If $\exists \lim_{x \rightarrow\infty}f(x)=L$ for some $L\in \mathbb{R}$, then the function $f$ must be bounded.

Proof: Let's assume that $f$ isn't bounded. Then in order to prove the statement above, this assumption must give us that $\nexists \lim_{x \rightarrow\infty}f(x)=L$.

If $f$ isn't bounded in $[a,\infty)$ then $\nexists C \in \mathbb{R} | f(x) \leq C, \forall x \in [a,\infty)$.

In other terms, $\forall N > 0, \exists \omega \geq a | \forall x\in [a,\infty), x > \omega \Rightarrow f(x) > N$

or

$\forall N < 0, \exists \omega \geq a | \forall x\in [a,\infty), x > \omega \Rightarrow f(x) < N$

But the statement above, is equivalent to the statement:

$\lim_{x \rightarrow\infty}f(x)=\infty$ and $\exists \lim_{x \rightarrow\infty}f(x)=-\infty$ respectively.

But then this is equivalent to $\nexists \lim_{x \rightarrow\infty}f(x)=L$

Hence, as we proved the contrapositive statement, the claim must hold true.

$\blacksquare$


I'd be glad if you could share some tips for improvements, and maybe share your own proofs, so we can discuss them together. Thank you!

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  • $\begingroup$ Consider $f(x)=x\sin(x)$. It is continuous, unbounded and oscillates. $\endgroup$ Sep 29, 2021 at 19:04
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    $\begingroup$ I don't think the statement that there is no $C$ such that $f(x) \le C$ for all $x$ immediately implies that there exists $\omega \ge a$ such that for all $x > \omega$ we have $f(x) > N$. Specifically, the "for all $x > \omega$" part seems like it would take more explanation. $\endgroup$ Sep 29, 2021 at 19:05
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    $\begingroup$ @LázaroAlbuquerque I don't think that $\lim_{k \rightarrow \infty} f(x)$ exists for that function. $\endgroup$ Sep 29, 2021 at 19:06
  • $\begingroup$ @user6247850 I thought that if it's not bounded, then it's certainly always increasing or decreasing. But then I realized that the comment above, had a very good counter example for my statement. However, I don't really know how I'd compensate for this in my proof. Do you got any tips. $\endgroup$
    – Tanamas
    Sep 29, 2021 at 19:09
  • $\begingroup$ @user6247850 Oh yes, that's true. I totally forget that's what I assumed in my claim already. Thanks. $\endgroup$
    – Tanamas
    Sep 29, 2021 at 19:09

2 Answers 2

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It seems that you did not consider oscillating and increasing magnitude.

We can find $\omega > a$ such that for $x > \omega$, $|f(x) - L| \le 1$, from here, you can get a bound of $f$ on $(\omega, \infty)$.

Also, there is a famous result that states that continuous function on compact set, $[a, \omega]$ attains its maximum and minimum. Combining these two portions, you should be able to show that it is bounded.

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The fourth line of the proof is incorrect. Think about an unbounded function that wiggles up and down.

Personally, I think a direct proof is more straightforward. Here is one such proof: Since $\lim_{x\rightarrow\infty} f(x) = L$, there exists an $N>0$ such that $f(x) \in (L-1, L+1)$ for all $x\ge N$. On the other hand, since $f$ is continuous and $[a,N]$ is compact, $f$ attains a maximum and minimum on $[a,N]$ (e.g., see this post). Therefore, we have that $$D := \min_{a\in [a,N]} f(x) > -\infty\quad\text{and} \quad U := \max_{x\in [a,N]} f(x)<\infty.$$ It follows that $\min\{L-1, D\}\le f(x) \le \max\{L+1, U\}$ for all $x\in [a,\infty)$. Hence, $f$ is bounded (from above and below).

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