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X tends to 0 Lim cos (1 / x), X tends to Pai / 2 TaNx
When x → 0, 1 / X →∞, cos (1 / x) has no limit, and the limit is not ∞
When x → π / 2 +, TaNx → - ∞, when x → π / 2 -, TaNx → + ∞
Z = Y / x, x = e ^ t, y = 1-e ^ 2T, find DZ / dt
z'x=-y/x^2,z'y=1/x
x't=e^t,y't=-2e^(2t)
dz/dt=z'x*x't+z'y*y't=-y/x^2*e^t+1/x*-2e^(2t)
Substitute the others yourself
Known SiNx + cosx = half minus radical three (0
sinx+cosx=(1-√3)/2
Given that the function f (x) = SiNx cosx and f '(x) = 2F (x), f' (x) is the derivative of F (x), then sin2x = () A. 1 three B. -3 five C. 3 five D. -1 three
∵ function f (x) = SiNx cosx and f '(x) = 2F (x),
∴cosx+sinx=2sinx-2cosx，
∴sinx=3cosx，
∴tanx=3，
∴sin2x=2sinxcosx=2sinxcosx
sin2x+cos2x=2tanx
tan2x+1=3
5．
Therefore, C
The known function (FX) = (SiNx cosx) sin2x / SiNx finds the monotonically increasing interval of function FX Why first 2K π - π / 2 ≤ 2x - π / 4
This is using the period of the function y = Sint itself, which is 2 π
Independent of known functions
Given the vector a = (SiNx, cosx), B = (1, - 2), and a ⊥ B, then tan2x=
Because a ⊥ B
So AB = 0
That is, sinx-2cosx = 0
sinx=2cosx
tanx=2
tan2x=2tanx/(1-tan^2(x）)=-4/3
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