Proof of Simple Limit: f(x)≤g(x) for all x

[dylanm]

Hi folks. I'm a longtime lurker who is starting to explore proof-based mathematics, and I'm having trouble figuring out what I can and cannot do in a problem. I'm stuck on this simple problem from the wonderful Spivak Calculus book:

If $$f(x)\le g(x) \forall x$$, then $$\lim_{x\to a}f(x)\le\lim_{x\to a}g(x)$$

Intuitively, this is obvious. But when I fiddle with it, taking $$g(x)-f(x)\ge0$$ as my function and proving that the limit of this function, $$c=\lim_{x\to a}g(x)-\lim_{x\to a}f(x)$$, is $$\ge0$$, I become stuck. No amount of algebra seems to give me a clean relation between zero and c. I know that one can say that c can be made arbitrarily close to a value of a function that must be $$\ge0$$, but I'm not sure if I'm allowed to use this sort of thinking in a chapter (5) that has just introduced the concept of a limit.

 

[LeonhardEuler]

Are you allowed to use that the difference of the limits is the limit of the difference? Even if not, it's not too difficult to prove and then use in the problem.

 

[Hurkyl]

Here's an important trick that you might find useful.

$$(\forall \epsilon \geq 0: 0 \leq x \leq \epsilon) \implies x = 0$$

To say it differently, if x is nonnegative, and every nonnegative number is at least as large as x, then x is zero.

Try proving this statement. Once you've accomplished that, add it to your toolbox, because it is an often useful little trick.