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


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

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.


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!

  • $\begingroup$ Consider $f(x)=x\sin(x)$. It is continuous, unbounded and oscillates. $\endgroup$ Sep 29, 2021 at 19:04
  • 2
    $\begingroup$ 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


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.


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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