Problem: Let $f:\mathbb{R}^+\rightarrow \mathbb{R}$ be bounded and decreasing. Then $lim_{x\rightarrow 0}f(x)$ exists.
My attempt:
Let $\epsilon>0$. Let $L=\sup_{x\in \mathbb{R}^+}f(x)$. Then there exists $x_0\in \mathbb{R}^+$ such that $f(x_0) > L-\epsilon$. Choose $\delta=x_0$. Then $|x|<x_0$ $\implies$ $L-\epsilon$ <$f(x_0)\leq f(x)$ $<L+\epsilon$. Hence, $|f(x)-L|<\epsilon$.
Is my attempt correct?
Yes, your attempt is correct, indeed $0 < x < x_0$ implies $f(x) \ge f(x_0)$ by decreasingness, and $f(x) \le L < L+\varepsilon$ by $L$ being the sup of all values, to be even more explicit. Part of the definition of the limit is also noting that $0$ is a limit point of $\Bbb R^+$, which is obvious, but could be mentioned too.