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2, b is wrong. For example, f(x)=sinx when x!=0, f(x)=1 When x=0, so that f(x) is continuous at x=0, and f(x)+f(-x)=0, so the limit = 0, but f(0)=1

A is correct, because the denominator is ->0, and the limit exists, only the numerator is also ->0, and f(x) is continuous, so the limit value = function value, there is f(0) = 0

C is correct, by the conclusion of A, F'(x) is the limit value in =c.

d is correct, because the limit exists in d, f(x) is continuous, and f(0) can be added or subtracted from the molecule, split into two limits, which illustrates f'(0) Exists.

3, C error. Because tanh-sinh is not infinitesimal of the same order as H2.

The functions and denominators in molecular parentheses in a, b, and d are infinitesimal of the same order. Therefore.

The first question, C is right, which is the definitive deformation of the derivative, which should be visible.

In the second question, according to the definition of derivatives, a and c are excluded first, and the closest thing to the definition is b

b is wrong.

The denominator of a tends to 0, and the limit exists, only the numerator also tends to 0, and f(x) is continuous, and there is f(0)=0

c is clearly correct.

The D numerator is combined with addition and subtraction f(0) and splits into two limits, illustrating f'(0) Exists.

z=x^3-y^3+3x^2+3y^2-9x=(x^3+3x^2-9x)-(y^3-3y^2)

dz=3(x^2+2x-3)dx-3(y^2-2y)dy

x^2+2x-3=0，y^2-2y=0

x=1 or x=-3) and (y=0 or y=2)

Stability points: (1,0), (3,0), (1,2), (3,2).

z1=x 3+3x 2-9x at x=-3 and minimum at x=1, z2=-(y 3-3y 2) at y=0 and maximum at x=2.

So the function z obtains a maximum value at point (-3,2) =(-27+27+27)-(8-12)=27+4=31

So the function z obtains a minimum at the point (1,0)=(1+3-9)-(0-0)=-5+0=-5

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First, the function defines the domain as x>0.

When 00, therefore (0,1) is the concave interval of the function;

When x>=1, y=lnx, y'=1/x, y''=-1 x 2<0, so (1,+ is the convex interval of the function.