# Thread: uniformly continuous and boundeness

1. ## uniformly continuous and boundeness

Let a be in R. Suppose f is a real valued function on [a, infinity) satisfying that lim as x -> infinity of f(x) is L, where L is in R.
I need to show:

f is bounded on [a, infinity)

here is what I have so far:
if f is bounded on [a, infinity), then there exists a constant M s.t. |f(x)| <= M for all x in [a, infinity)

From the fact that lim f(x) = L, I can say that given epsilon > 0, there is a delta s.t |f(x) - L| < epsilon.

if L=0, then |f(x)| < epsilon. I can set M <= to epsilon, but I don't think this last part makes any sense because there is no way I can claim that L =0

or what if I say that since |f(x) - L| < epsilon, then |f(x) - L + L| < epsilon + |L|. If I let M = epsilon + |L|, then it would work, wouldn't it?
Please provide me with some hints

By the way, the second part of the problem is to show f is uniformly continuous on [a, infinity)

2. I assume you're given that f is continuous?

Edit: Or you have to show existence of an $a$ such that f is bounded on [a,inf)? Use the definition of a limit at infinity, i.e. there is an X such that for x>X, |f(x)-L|<epsilon.

3. yes, it is a continuous function

4. Okay, find some b>a (as above) so that f is bounded on [b,inf). And f is bounded on [a,b] because it's continuous on a compact set.

5. how do I prove that f is bounded on the [a,b]. I understand that if I show f is bounded on [a, b], since it is closed, then it is compact. the f is continuous on the compact set.

6. Originally Posted by inthequestofproofs
how do I prove that f is bounded on the [a,b]. I understand that if I show f is bounded on [a, b], since it is closed, then it is compact. the f is continuous on the compact set.
Closed intervals in R are compact. Continuous functions are bounded on compact sets. Since f is continuous, f is bounded on [a,b].