I have to prove that if $(x_1 \gt x_2) \Rightarrow (f(x_1) \ge f(x_2))$, then $$\forall a \in \mathbb R \exists L \in \mathbb R \lim_{x \to a^+}f(x) = L$$
I have a feeling that L = $inf_{x \in (a,a+ \delta)}f(x) = f(a)$, but I can't prove that for each $\epsilon$ I'll find $x \in (a,a+ \delta)$, $|f(x)-f(a)|\lt \epsilon$. Is my guess wrong? How should I go?
On
The feeling is in the right track, about looking at $\inf$s and $\sup$s. Here we will use $\sup$, $\inf$ will be used to prove the analogous result for decreasing functions. You can't have $\inf_{(a,a+\delta)}$ because you want fo find $\delta$! Look at the set: $$\{ f(x) \mid x < a\}. $$The set is bounded by $f(a)$, so we have the existance of $L = \sup\{f(x) \mid x < a\}$. To check that this works, let $\epsilon > 0$ be any real number. By the definition of $\sup$, there exists $x_0 < a$ such that: $$L-\epsilon < f(x_0) \leq L$$We don't actually need the $f(x_0) \leq L$ part, I just put it there to help you see the bigger picture. Note that the above gives $L-f(x_0) < \epsilon$. Recall that given $\epsilon$, we want $\delta$. Pick $\delta = a - x_0 > 0$. Now we check the definition of the lateral limit. Let $x \in \Bbb R$ such that $a-\delta < x < a$. Note that $a-\delta < x \implies x_0 < x$, so that $f(x_0) \leq f(x)$. So: $$|f(x) - L| = L-f(x) \leq L - f(x_0) < \epsilon,$$and we're done. We have $|f(x) - L| = L-f(x)$ because $x < a$ and hence $f(x) < L$.
Let $L=\inf_{x> a} f(x)$.
Then given $\epsilon > 0$, we can find some $x_0>a$ such that $f(x_0)<L+\epsilon$ (because $L=\inf_{x>a}f(x)$). Then whenever $x\in (a,x_0)$, we know that $L \leq f(x) \leq f(x_0)<L+\epsilon$, and thus $|f(x)-L|<\epsilon$. Letting $\delta = x_0-a$, we see that $x \in (a,a+\delta)$ implies that $|f(x_0)-L|<\epsilon$. Thus $\lim_{x \to a^+} f(x)=L$.