Why is the proposition "when limx →∞ f (x) = 0, there is x > 0, when x > x, f (x) is bounded" wrong?

Why is the proposition "when limx →∞ f (x) = 0, there is x > 0, when x > x, f (x) is bounded" wrong?

When x tends to + ∞, limf (x) = 0, expressed as, according to the limit definition, for any ε> 0, there is x > 0, so that when x > x, |f (x)|< ε. From the last inequality, it seems that f (x) is bounded, but in fact, it is not, because there is a difference here ε Is arbitrary, and X is dependent on ε Yes, i.e. selected ε Different, then the found x is also different, so x is a changing number, not a predetermined number, but the function is bounded in a certain interval, which is required to be determined. It is meaningless to say that the function is bounded in a changing interval. This problem reflects the local boundedness of the function. Note that this local boundedness can only ensure that the function is bounded in a certain range, But this range cannot be given accurately

F (0) = f (1) = 0, f (1 / 2) = 1. It is proved that there is at least one point in (0,1) so that f '(x) = 1 F (x) is continuous on [0,1] and differentiable in (0,1).

According to Lagrange's theorem, there exists ξ 1 ∈ (0,1 / 2), satisfying f '（ ξ 1) = [f (1 / 2) - f (0)] / (1 / 2-0), i.e. f '（ ξ 1)=2①； Existence of the same reason ξ 2 ∈ (1 / 2,1), satisfying f '（ ξ 2) = [f (1) - f (1 / 2)] / (1-1 / 2), i.e. f '（ ξ 2)=-2②； Consider the limit LIM (△ x → 0) f '(x + △ x). Since f (x) is differentiable in (0,1), that is, f' (x) exists, LIM (△ x → 0) f '(x + △ x) = f' (x), that is, f '(x) is continuous in (0,1), so there is at least one point x ∈（ ξ 1, ξ 2) , i.e. x ∈ (0,1), satisfying f '（ ξ 1)=2＞f'(x)=1＞f'( ξ 2)=-2.

F (x) is a function defined on R, X belongs to R, f (x) = f (x + 1) + F (x-1) is constant. It is proved that FX is a periodic function. If f (3) = 2, find f (2010)

f(x+1)=f(x)-f(x-1)
Then f (x) = f (x-1) - f (X-2)
So f (x + 1) = - f (X-2)
So f (x + 1) = - f (X-2) = f (X-5)
f(x+1)=f(x-5)
F (x) is a periodic function, t = 6
f(2010)=f(335*6)=f(6)
f(6)=-f(3)=-2

The function f (x) defined on R satisfies: for any a, B ∈ R, there is always f (a + b) - [f (a) + F (b)] = 2010, then the following statement is correct A. F (x) - 1 is an odd function B, f (x) + 1 is an odd function C, f (x) - 2010 is an odd function D, and f (x) + 2010 is an odd function

The function f (x) defined on R satisfies that for any a, B ∈ R, there is always f (a + b) - [f (a) + F (b)] = 2010
Let a = b = 0 get f (0 + 0) - [f (0) + F (0)] = 2010
So f (0) = - 2010
So f (0) + 2010 = 0
Because the odd function defined on R must pass through the origin
Therefore, D can be selected by the exclusion method
(ABC option)

The function f (x) defined on R is an odd function and satisfies f (x + 6) = f (x). If f (1) = 2010, the value of F (2009) + F (2010) is equal to

Because f (x + 6) = f (x), f (x) is a periodic function with a period of 6. F (2009) = f (334 × 6 + 5) = f (5) = f (- 1) = - f (1) = - 2010 [odd functions have f (- x) = - f (x)] f (2010) = f (335 × 6 + 0) = f (0) = 0 [odd function f (0) = 0] f (2009) + F (2010) = - 2010 happy study

If the function f (x) defined on R satisfies f (x) = 3 ^ (x-1), x < = 0, f (x-1) - f (X-2), x > 0, then f (2010) =___

Because f (2010) = f (2009) - f (2008) = f (2008) - f (2007) - f (2008) = - f (2007) = f (2004)
Therefore, the period T = 6, that is, f (2010) = f (0) = 1 / 3