This is about Chapter 8 Question 8 in Spivak:
Suppose that $f$ is a function such that $f(a) \leq f(b)$ whenever $a < b$. Prove that $\displaystyle\lim_{x\to a^{-}}f(x)$ and $\displaystyle \lim_{x\to a^{+}}f(x)$ both exists.
Here is my work so far for $\displaystyle\lim_{x\to a^{-}}f(x)$:
Considering the interval $(-\infty, a)$. Since $f(a) \leq f(b)$ whenever $a < b$ it is clear that for every $x\in (-\infty, a)$ $f(x) \leq f(a)$. That is $f(x)$ is bounded above at $(-\infty, a)$ by $f(a)$. Since $f(x)$ is bounded above, $f(x)$ has a least upper bound, let it be $\alpha$.
I want to prove that
$\displaystyle\lim_{x\to a^{-}}f(x) = \alpha$
By definition $\forall \varepsilon > 0, \exists \delta > 0$ st. $\forall x$ if $ 0 < a-x< \delta$ then $|f(x) - \alpha | < \varepsilon$
How would I choose $\delta$ in this case?
Thanks.