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As it appears, you seem to be seeking the limit of sin(4x)*sin(6x) as x tends to 0.
Both components of the product tend to 0 as x tens to 0 and so the limit is 0. Bit I suspect that is not the limit that you are looking for.
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What is the integral of sin squared 2x?
You would probably use a power-reduction trig identity to solve this equation. This states that sin2(x) = (1 - cos(2x))/2 Therefore, sin2(2x) = (1 - cos(4x))/2, or (1/2)(1 - cos(4x)) So, ∫ (1/2)(1 - cos(4x)) dx = (1/2) ∫ (1 - cos(4x)) dx. Then, ∫ (1-cos4x)dx = x - (1/4)sin(4x) + c Now, multiply that by (1/2) to get: (x - (sin(4x)/4) + c)/2 Since c is an arbitrary constant, we have: ½(x - sin(4x) / 4) + c OR 1/8 * (4x - sin(4x)) + c
If f(x) cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then find the limit of 1 - f(x)35(sinx)2 as x tends to 0 (zero).?
I'm sorry the question is not correctly displayed. If f(x) = cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then, find the limit of {1 - [f(x)]^3}/[5(sinx)^2] as x tends to 0 (zero).
What happens to the value of y when the value of x changes in y equals 4x?
y = 100 - 4x ( a ) Find the value of y when x = 20.
What is -4x divide by 1?
-5
8x plus 30 equals 4x plus 18 solve for x?
8x+30=4x+18 -4x -4x 4x+30=18 -30 -30 4x=-12 4x/4=-12/4 x=-3
