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Limx = 0, (TaNx SiNx) / SiNx to the third power. What's wrong with me? The third power of (TaNx SiNx) / SiNx = (x-x) / x ^ 3 But that's not right. What's wrong with me?
Don't replace with equivalent infinitesimal when adding and subtracting
It's easy to make mistakes
You're wrong here
Original formula = LIM (SiNx / cosx SiNx) / (SiNx) ³
=lim(1/cosx-1)/(sinx) ²
=lim(1-cosx)/cosx(sinx) ²
Because 1-cosx~x ²/ two
So the original formula = LIM (x ²/ 2)/(x ² cosx)=1/2
Find limx - 《 0 (1-sinx) Cotx power Cotx is the symbol of the square Please list the equations,
(1-sinx) the power of Cotx is written as (1-sinx) ^ (Cotx), then the original formula = (1-sinx) ^ (cosx / SiNx), here I omit = {[1 + 1 / (1 / - SiNx)] ^ (1 / - SiNx)} ^ (- cosx) note that the important limit (1 + 1 / x) ^ x tends to e, when x tends to infinity, and when 1 / - SiNx tends to infinity, X - tends to 0, so Lim [
If the x power of 10 = 2 and the Y power of 10 = 9, find the 2x power of 100 - the Y power of 1 / 2
Original formula = 100 ^ [2x - (1 / 2) ^ y]
=100^2x÷100^(1/2)^y
=(10^x)^4÷10^y
=2^4÷9
=16/9
Given that the x power of 10 = 1 / 2 and the Y power of 10 = 2, find the value of 2x + y power of 100
The power of 2x + y of 100 is equal to the power of 4x + 2Y of 10
It is known that the x power of 10 is equal to one-half, so the 4x power of 10 is equal to 1 / 2 and the fourth power is 1 / 16
The Y power of 10 is equal to 2, and the 2Y power of 10 is equal to 4, so the answer is 1 / 16 multiplied by 4 is 1 / 4
Graph area enclosed by y = x cubic and y = 2x Please help me thank you
The simultaneous intersection of y = x ^ 3 and y = 2x is x ^ 2 = 2, so the intersection coordinates are (2 ^ 1 / 2,2 * 2 ^ 1 / 2) and (- 2 ^ 1 / 2, - 2 * 2 ^ 1 / 2). In fact, the graph of the Y axis for two days is symmetrical according to the origin. Just calculate the case of x > 0 and multiply the answer by 2 to integrate (2x-x ^ 3) (0 to 2 ^ 1 / 2) = (x ^ 2-1 / 4 * x ^ 4) (0 to 2 ^ 1 / 2) = 1, so
Find the area of the graph enclosed by the cubic power of curve y = x and the square of y = 2x-x?
Simultaneous y = x ^ 2, y = x ^ 3, the solution is: x = 0, x = 1,
Closed graph area = ∫ upper 1 lower 0 (x ^ 2-x ^ 3) DX = (x ^ 3 / 3-x ^ 4 / 4) | upper 1 lower 0 = (1 / 3-1 / 4) - 0 = 1 / 12
The application of definite integral in finding the area of plane graph
It's actually very simple, just a set of formulas
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