# Proof that if a function has a limit for large x, then the function is bounded.

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!

• Consider $f(x)=x\sin(x)$. It is continuous, unbounded and oscillates. Sep 29, 2021 at 19:04
• 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. Sep 29, 2021 at 19:05
• @LázaroAlbuquerque I don't think that $\lim_{k \rightarrow \infty} f(x)$ exists for that function. Sep 29, 2021 at 19:06
• @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. Sep 29, 2021 at 19:09
• @user6247850 Oh yes, that's true. I totally forget that's what I assumed in my claim already. Thanks. Sep 29, 2021 at 19:09

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.
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).