To find the limit LIM (x → 0) (x ^ 3sin1 / x) / (1-cos ^ 2x), the detailed process ～

To find the limit LIM (x → 0) (x ^ 3sin1 / x) / (1-cos ^ 2x), the detailed process ～

Molecule: (1-cos ^ 2x) = (1 + cosx) (1-cosx)
1-cosx~0.5x^2
By substituting, we can get the original formula = LIM (x → 0) (2xsin1 / x) / (1 + cosx)
Substituting the numerator and denominator into x = 0, the denominator equals 2, and the numerator equals 0 (sin1 / 0 = sininfinity = is an uncertain constant, multiplying by zero must be zero)
The limit is 0

Zu Chongzhi is the first person in the world to get the value of pi to 6 decimal places. This achievement is at least several years earlier than that of foreign countries

Zu Chongzhi's outstanding achievement in mathematics is about the calculation of PI. Before the Qin and Han Dynasties, people used "track one week" as pi, which is called "ancient rate". Later, it was found that the error of ancient rate was too large. PI should be "circle diameter one and more than three days", but there were different opinions about how much more. Until the Three Kingdoms period, Liu Hui proposed the calculation of circle

Calculate the surface area and volume of the figure below
A cuboid is 30 decimeters long, and a cylinder is dug on its long and high side,

Surface area = surface area of cuboid + side area of cylinder - bottom area of cylinder × 2
=(30×5+30×20+20×5)×2+3.14×10×5-3.14×10²÷4×2=1935.5 DM²
Volume = volume of cuboid - volume of cylinder
=30×20×5-3.14×10²÷4×5=2607.5 DM³

The x power of F (E) = the square of X - 2x + 3, X belongs to the analytic formula and definition field of [2,3] for finding f (x),

If x belongs to [2,3], then the power of e ^ x belongs to [e ^ 2, e ^ 3]. Let x = LNT of T = e ^ X. then f (T) = (LNT) ^ 2-2lnt + 3. So f (x) = (LNX) ^ 2-2lnx + 3 is defined as [e ^ 2, e ^ 3]

Because a is divisible by B, a is a multiple and B is a factor

∵b|a
A is a multiple of B,
B is a divisor of A

Is the length 2a of the major axis of the ellipse the sum of the distances 2a from a point on the ellipse to two focal points the same?

Yes,
It's the same. It's all 2A

Why are prime numbers within 20 2,3,5,7,9,11,13,17,19

It can only be divisible by 1 and itself, and 1 is neither prime nor composite

Given the intersection (2,0) of quadratic function image and X axis (- 1,0), and the intersection (0, - 1) of quadratic function image and Y axis

Let the analytic formula of quadratic function be y = a (X-2) (x + 1), substitute the point (0, - 1), get - 1 = a (0-2) (0 + 1), get a = 12; so the analytic formula of quadratic function is y = 12 (X-2) (x + 1); y = 12 (X-2) (x + 1) = 12 (X-12) 2-98, vertex coordinates are (12, - 98)

Let the image of quadratic function f (x) = ax ^ 2 + BX + C take the y-axis as the symmetry axis, and a + B = 1, e is known, and if the point (x, y) is on the image of y = f (x)
Then point (x, y ^ 2 + 1) can find the analytic expression of G (x) on the image of function g (x) = f (f (x))

(x) The image of = ax ^ 2 + BX + C takes the Y axis as the symmetry axis - B / (2a) = 0b = 0A + B = 1, a = 1F (x) = ax ^ 2 + BX + C = x ^ 2 + CG (x) = f (f (x)) = (x ^ 2 + C) ^ 2 + C (x, y ^ 2 + 1) on the image of function g (x) = f (f (x)) (x ^ 2 + C) ^ 2 + C = (x ^ 2 + C) ^ 2 + 1C = 1g (x) = (x ^ 2 + 1) ^ 2 + 1 = x ^ 4 + 2x ^ 2 + 2

Given x > y > z > 1, then the integer solution suitable for the equation XYZ + XY + YZ + ZX + X + y + Z = 2003 is___ ．

The original formula = XY (Z + 1) + Z (x + y) + X + y + Z = XY (Z + 1) + (Z + 1) (x + y) + (Z + 1) - 1, = (XY + X + y + 1) (Z + 1) - 1, = (x + 1) (y + 1) (Z + 1) - 1, that is: (x + 1) (y + 1) (Z + 1) = 20042004 = 2 × 2 × 3 × 167, then 2004 is obtained by multiplying three numbers, and the minimum Z is 2, Z + 1 ＞ = 3. Then it can only be 3 × 4 × 167. Because x ＞ y ＞ Z ＞ 1, so x = 166, y = 3, z = 2 So the answer is: x = 166, y = 3, z = 2