what does this expression mean when talkin about calculus?

Question

Suppose that f(x) is defined for all x in
U= (4,5) u (5,6) <---- i dont quite what this is talking about
and that lim f(x)= L
x->5

why would these statements be not true?
I) if L >0 , then f(x) > o on U
III) If L= 0, then f(x)= 0 on U

but this would be true :
II) If f(x) > 0 on U, then L >/ 0

Answer 1

(4, 5) ∪ (5, 6)
means all points that are in either (4, 5) or (5, 6) - in other words, it is the interval (4, 6) without the point 5.

(I) and (III) should be pretty easy to find counterexamples for; e.g. for (I) suppose L = 2, and let f(x) = 3x - 13. Then lim(x->5) f(x) = 2 but f(x) < 0 on (4, 13/3) which is part of U.

The point is that the limit only tells us about the behaviour of the function as we approach 5, and the behaviour in other parts of U may be different. Similarly, for (III) we can just take f(x) = x-5. Then the limit is 0 as required, but f(x) ≠0 for all x in U.

For (II), the easiest way to prove this is probably by contradiction: suppose L < 0, and pick ε < |L|. Then we know there's a δ > 0 such that if 0 < |x-5| < δ then |f(x) - L| < ε (this is the definition of the limit). But we know f(x) > 0 and L < 0, so |f(x) - L| = f(x) + |L| > |L| > ε for any x in U. So we have a contradiction, so we must have L ≥ 0.

(Note for hawkeye3772: for II, if the function is greater than 0 on U it is only true that the limit must be greater than or equal to 0; the limit does not have to be greater than 0. For instance, consider f(x) = |x-5| which is strictly positive on U but has limit 0.)

Answer 2

U is the open set (4,6) /0 expressed as a union of open sets (4,5) u (5,6). Notice that neither (4,5) nor (5,6) actually 'contain' 5.
Lim x -> 5 f(x) = L simply says that as x approaches 5, the value of f(x) approaches L.

As for L > 0 => f(x) > 0 on U and L = 0 => f(x) = 0 on U, It's bogus because L is defined at a single point ( x = 5 in the limit) but that says -nothing- about the behavior of the function for other values of x over U.
However..... f(x) > 0 on U says that for every x on U, f(x) is greater than 0. And since L is a limit as x -> 5, f(x) must be > 0 at all of those points. Which means that f(x) > 0 at all x then L > 0 since L is a limit of f(x).
Limits are something that you have to look at and think about a while before they start to make sense ☺

Doug

Answer 3

U is the interval the function is defined on. So for this one, the function goes from x = 4 to x = 5 and from x = 5 to x = 6 (but not necessarily at x = 5).

I is not true; it is saying that if the limit is greater than zero, then the entire function must be greater than zero on the interval.

II is true; if the entire function is greater than zero, then the limit must be greater than zero.

II is not true; if the limit is zero, the entire function does not have to equal zero.