Real Analysis Proof?

Question

Prove theorem 2.5:

Suppose f : D (arrow) R and g : D (arrow) R, x is an accumulation point of D, and f and g have limits at x. If f(x) is less than or equal to g(x) for all x E D, then

lim f(x) less than or equal to lim g(x)

Answer 1

That uses 'x' a couple of times, so just to make things clearer, I'll call the point we're considering 'y'.

Lets call the limits of f and g at y A and B respectively.
Now, by definition of a limit:
For all ε > 0, there exists δ1 > 0 such that for all x with 0<|x-y|<δ1, |f(x)-A|<ε.
Similarly,
For all ε > 0, there exists δ2 > 0 such that for all x with 0<|x-y|<δ2, |g(x)-B|<ε.

Now, lets suppose (for the sake of contradiction) that B Now, from above there exists δ1 > 0 so that if |x-y|<δ1 then |f(x)-A|<ε, and δ2 > 0 so that if |x-y|<δ2 then |g(x)-B|<ε.

Lets pick x close enough to y so that both those conditions hold - eg x = y+min(δ1,δ2). Then the following holds:
g(x) < B+ε = B+(A-B)/2 = A-(A-B)/2 = A-ε < f(x). Contradiction.

The key idea is that if B>A, then there is a 'gap' between B and A - thus you just need to get smaller than the gap, and you'll get a contradiction.