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Function limit: the X of sin2 (x-1) / X-1 tends to the limit of 1, and the 2 after sin is the square
t=x-1
sin2t/t=2sintcost/t
t==>0,c0st==>1
sint/t ==1
The limit is 2
When x tends to zero, find the limit of sin squared minus x squared
When (x → 0), the original limit = 1 / (SiNx) ^ 2-1 / x ^ 2 = [x ^ 2 - (SiNx) ^ 2] / x ^ 4 = (2x-sin2x) / (4x ^ 3) = (2-2cos2x) / 12x ^ 2 (using lobita's law twice) = 2 * (2x ^ 2) / (12x ^ 2) {(x → 0), 1-cos2x is equivalent to (2x) ^ 2 / 2 = 2x ^ 2} = 1 / 3
Let f (x) = sin2x sin (2x - π 2). (1) find the maximum and minimum of F (x); (2) the opposite sides of a, B, C of △ ABC are a, B, C, C = 3, f (C2) = 14, if SINB = 2sina, find the area of △ ABC
(1) F (x) = sin2x sin (2x - π 2) = 1 − cos2x2 + cos2x = 12cos2x + 12 when cos2x = 1, the function gets the maximum value of 1; when cos2x = - 1, the function gets the minimum value of 0. (2) ∵ f (C2) = 14, ∵ 12cosc + 12 = 14, that is, COSC = - 12
Limx → infinite (2x-3) ^ 2 (3x + 1) ^ 3 / (2x + 1) ^ 5
In the arithmetic sequence {an}, if A2 + a3 + A10 + a11 = 36, then A5 + A8=______ ．
∵ in the arithmetic sequence {an}, A2 + a3 + A10 + a11 = 36 is known. According to the properties of arithmetic sequence, A5 + A8 = A3 + A10 = A2 + a11 = 36 △ 2 = 18 can be obtained, so the answer is 18
Given the arithmetic sequence an, A5 = A3 + A8, then the value of A10 is?
a1+4d=a1+2d+a1+7d
So A1 = - 5D
Obviously, this can't find the value of a10
Insufficient conditions
Given the arithmetic sequence {an} A6 = 5 A3 + A8 = 5, find A9 =?
A3+A8=A6+A5=5
A6=5
A5 = 0 arithmetic sequence {d = 5-0 = 5
a9=A6+3D
=5+15=20
In the arithmetic sequence {an}, A3 + A7 = 37, then A2 + A4 + A6 + A8=______ ．
In the arithmetic sequence {an}, A3 + A7 = 37, ∵ A3 + A7 = A2 + A8 = A4 + A6 = 37 ∵ A2 + A4 + A6 + A8 = 37 + 37 = 74, so the answer is: 74
In the arithmetic sequence an, if A3 + a11 = 10, then A6 + A7 + A8 is equal to 15
a3+a11=2a7=10
a7=5
a6+a8=2a7
a6+a7+a8=15
In the arithmetic sequence {an}, if A3 + A8 = 10, then 3A5 + A7=______ ．
From the properties of arithmetic sequence, we can get: 3A5 + A7 = 2a5 + (A5 + A7) = 2a5 + (2A6) = 2 (A5 + A6) = 2 (A3 + A8) = 20, so the answer is: 20
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