Prove that if f(x) is nondecreasing and bounded above on (a,b) then Lim x->a (f(x)) = M where M = least upper bound {f(x)|x \in (a,b)}.

I know that since f is bounded above, there is a number M so that f(x) \leq M for all x in (a,b). Then if I let L = least upper bound of f(x), I claim that L = lim x->a^- (f(x)). So, let \epsilon > 0. Then L - \epsilon < L. So L - \epsilon is not an upper bound for {f(x)}. So, for some x_0 \in (a,b), f(x_0) > L - \epsilon.

Now I am not sure how to choose \delta... I am thinking \delta might be b - x_0 but I am not sure.