$f(x)$ bounded above, continuous, and increasing on $(0, \infty)$ implies $\lim_{x \to \infty} f(x)$ exists. Suppose $f(x)$ is continuous, bounded, and increasing on $(0, \infty).$ Then $\lim\limits_{x \to \infty} f(x)$ exists. 
Here's something I came up with after browsing some related problems.
Let $A= \{f(t) : t \in (0 , \infty) \}$. Since $f$ is bounded, $\alpha = \sup A$ exists. The goal is to show $\alpha$ is the limit. Let $\epsilon > 0.$ Then $\alpha - \epsilon$ is not an upper bound for $A$. So there is some $c \in (0, \infty)$ such that 
$$
\alpha - \epsilon < f(c) \leq \alpha.
$$
If $x > c > 0$ then $f(x) \geq f(c)$ which gives 
$$
\alpha - \epsilon < f(c) \leq f(x) \leq \alpha < \alpha + \epsilon
$$
and 
$$
-\epsilon <f(x) - \alpha < \epsilon \implies |f(x) - \alpha| < \epsilon.
$$
Some questions: 


*

*If this argument is correct, where is continuity used, or have I not used it appropriately? 

*My initial approach involved sequences, but I couldn't really work out the details. Is there a sequential argument to be made?

 A: Your argument does not use the continuity of $f$. In fact, continuity is not necessary: what we actually need is $f$ is increasing and bounded.
Sequential argument is not to much different from your argument: take any increasing sequence $a_n$ diverges to infinity and take the limit $L=\lim f(a_n)$. You can see that the limit exists.
We need to show that $L$ is also the limit of $f(b_n)$, for any increasing $b_n$ which diverges to the infinity.
We can see that for each $n$, $f(a_n)\le f(b_m)$ for all but finitely many $m$. Therefore $f(a_n)\le \lim f(b_m)$ and thus $L\le\lim f(b_m)$. We can show the opposite inequality by the same argument.
A: $2$: The sequence $(f(n))_{n\in\mathbb{N}}$ is increasing and bounded and therefore convergent. Let $L$ be the limit.
$$\forall\epsilon>0\exists N\in\mathbb{N}\forall n\geq N:|f(n)-L|<\epsilon$$
Thus for any $x\in\mathbb{R},x>N$ we get $|f(x)-L|<\epsilon$

$1$: Our proofs didn't need $f$ to be continuous so $f$ being bounded and increasing is sufficient (like Hanul Jeon already pointed out)
