# Limit of nondecreasing function that is bounded above

• Sep 1st 2010, 12:57 PM
tarheelborn
Limit of nondecreasing function that is bounded above
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.
• Sep 1st 2010, 01:02 PM
HallsofIvy
Nothing wrong with $\displaystyle \delta= |a- x_0|$!
• Sep 1st 2010, 01:20 PM
tarheelborn
OK, suppose I choose \delta to be a - x_0 (since L = the limit as x -> a from the left). Now I must show that if 0 < a - x_0 < \delta, then |f(x) - L| < \epsilon. Now a - x < \delta \implies a - x < a - x_0, so x > x_0. Since f is nondecreasing, f(x) > f(x_0) > L - \epsilon. Since L = l.u.b.{f(x)}, then f(x) \leq L. Then
L - \epsilon \leq f(x) \leq L
L - \epsilon \leq f(x) \leq L \leq L + \epsilon
L - \epsilon \leq f(x) \leq L + \epsilon
-\epsilon \leq f(x) - L \leq \epsilon
|f(x) - L| < \epsilon QED

Right?
• Sep 1st 2010, 01:22 PM
Plato
Quote:

Originally Posted by tarheelborn
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)}.

It is not true. I think you mean $\displaystyle \displaystyle \lim _{x \to b} f(x) = M$.
But you cannot even prove that!

This is true, $\displaystyle \displaystyle \lim _{x \to b^{-}} f(x) = M$.
• Sep 1st 2010, 01:49 PM
tarheelborn
Oops, duh...

So do I need to change my \delta to relate to b or can I still use a - x_0?
• Sep 1st 2010, 02:03 PM
Plato
If $\displaystyle \varepsilon > 0$ then $\displaystyle M - \varepsilon < M = \sup \left\{ {f(x):x \in (a,b)} \right\}$.

So $\displaystyle \left( {\exists x_0 \in (a,b)} \right)\left[ {M - \varepsilon < f(x_0) \leqslant M \leqslant f(b)} \right]$.

Now finish.

Now finish.
• Sep 1st 2010, 05:31 PM
tarheelborn
I have to have an epsilon/delta proof. So I think I got goofed by trying to follow the example from the limit as x approaches a from the right. I should have been dealing with b all along. So, if \epsilon > 0, then L - \epsilon < L. So L - \epsilon is not an upper bound... So for some x_0 \in (a,b), f(x_0) > L - \epsilon. Choose \delta to be b - x_0. Now I must show that if 0 < b - x < \delta, then |f(x) - L| < \epsilon.
b - x < \delta ==> b - x > b - x_0 so x > x_0. Since f is nondecreasing,
L - \epsilon \leq f(x_0) \leq f(x). Since L = l.u.b.{f(x)}, certainly f(x) \leq L. Now I have
L - \epsilon \leq f(x) \leq L
L - \epsilon \leq f(x) \leq L \leq L + \epsilon
-\epsilon \leq f(x) - L \leq \epsilon
|f(x) - L| < \epsilon QED

Does that get it? Thank you for your help!