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!
Best Answer
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.