Uniform Continuity on Closed and Bounded Intervals

[bluskies]

Homework Statement

Suppose that $$f: [0, \infty) \rightarrow \mathbb{R}$$ is continuous and that there is an $$L \in \mathbb{R}$$ such that $$f(x) \rightarrow L$$ as $$x \rightarrow \infty$$. Prove that f is uniformly continuous on $$[0,\infty)$$.

2. Relevant theorems

If $$f:I \rightarrow \mathbb{R}$$ is continuous on I, where I is a closed, bounded interval, then f is uniformly continuous on I.

Or, the definition of uniform continuity:

A function $$f:E \rightarrow \mathbb{R}$$ is uniformly continuous on E iff for every $$\epsilon >0$$, there is a $$\delta >0$$ such that $$|x-a|<\delta$$ and $$x,a \in E$$ imply $$|f(x)-f(a)|<\epsilon$$.

Archimedean Property: For every $$x \in \bb{R},$$ there exists an $$n_x$$ such that $$x<n_x$$.

The Attempt at a Solution

I was trying to use the theorem I listed, since we already know that f is continuous on the interval and that the interval $$[0, \infty)$$ is closed, as well as bounded below. My main question is whether the interval is bounded above. From the Archimedean Property, we can see that the real line is bounded above. Does the idea that infinity is bounded above follow directly, or is there something that I'm missing?Any help would be very much appreciated. Thank you!Edit: I'm sorry about the latex attempt, I'm still trying to figure out how to use it on the forums...the R is supposed to be the real numbers.

 

[Mark44]

Homework Statement

Suppose that $$f: [0, \infty) \rightarrow \bb{R}$$ is continuous and that there is an $L \in \bb{R}$ such that $f(x) \rightarrow L$ as $x \rightarrow \infty$. Prove that $f$ is uniformly continuous on $[0,\infty)$.

2. Relevant theorems

If $f:I \rightarrow \bb{R}$ is continuous on $I$, where $I$ is a closed, bounded interval, then $f$ is uniformly continuous on $I$.

Or, the definition of uniform continuity:

A function $f:E \right \bb{R}$ is uniformly continuous on $E$ iff for every $\epsilon >0$, there is a $\delta >0$ such that $|x-a|<\delta$ and $x,a \in E$ imply $|f(x)-f(a)|<\epsilon$.

Archimedean Property: For every $x \in \bb{R},$ there exists an $n_x$ such that $x<n_x$.

The Attempt at a Solution

I was trying to use the theorem I listed, since we already know that f is continuous on the interval and that the interval $[0, \infty)$ is closed, as well as bounded below. My main question is whether the interval is bounded above. From the Archimedean Property, we can see that the real line is bounded above. Does the idea that infinity it bounded above follow directly, or is there something that I'm missing?

Any help would be very much appreciated. Thank you!

Edit: I'm sorry about the latex attempt, I'm still trying to figure out how to use it on the forums...

Surround your expressions with [ tex] and [ /tex] tags, without the leading spaces. For example,
$$f: [0, \infty) \rightarrow \bb{R}$$

 

[bluskies]

Thank you for your help - I think I've fixed the Latex problems. Do you think I'm going in the right direction on the homework problem?

 

[Mark44]

The interval [0, infinity) is NOT bounded above. It makes no sense to say that infinity is bounded above.