To find the indefinite integral of SiNx \ (cosx) to the fourth power

To find the indefinite integral of SiNx \ (cosx) to the fourth power

The original formula = - ∫ D (cosx) / (cosx) ^ 4 = 1 / 3 * 1 / (cosx) ^ 3 + C

Find the indefinite integral of SiNx power of E!

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0

sin²xcos²x
=(1/4)(2sinxcosx)²
=sin²(2x)/4
=[1-cos(2x)]/8
=1/8 -cos(2x)/8
∫(sin²xcos²x)dx
=∫[1/8 -cos(2x)/8]dx
=x/8 -sin(2x)/16 +C

Find the limit LIM (x tends to 0) [SiNx / x] ^ (1 / x ^ 2) Urgent, as the title

lim[sinx/x]^(1/x²)
x→0
=lim[(x+sinx-x)/x]^(1/x²)
x→0
=lim[1+(sinx-x)/x]^{[(x/sinx-x)(sinx-x)/x](1/x²)}
x→0
=lim e^{[(sinx-x)/x](1/x²)}
x→0
=lim e^[(sinx-x)/x³]
x→0
=lim e^[(cosx-1)/3x²]
x→0
=lim e^[-sinx/6x]
x→0
=e^(-1/6)

LIM (x tends to 0): 5x + (SiNx) ^ 2-2x ^ 3 / TaNx + 4x ^ 2 can we replace the individual numbers of the numerator and denominator with equivalent infinitesimal

Certainly.
In this case, the lowest order infinitesimal of the molecule is 5x, and the denominator is TaNx
So just ask for the limit of 5x / TaNx

Lim x-0 (SiNx) ^ 2 / 1-cosx =? Substitute with equivalent infinitesimal Limx ^ 2 / x = limx = 0 (x-0) but the answer is 2, which is also done with equivalent infinitesimal. Why can't I do that? The answer is not right? The solution..... X-0 means that x tends to 0

Classmate 1-cosx is equivalent to 0.5x ^ 2
Original formula = limx ^ 2 / 0.5x ^ 2
=lim1/0.5=2

The limit of LIM (SiNx ^ m / (TaNx) ^ n) at x → 0 is solved by the property of equivalent infinitesimal!

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0

The limit of LIM (TaNx SiNx / x) is - 1
What you might want to calculate is
lim(tanx-sinx）/x=limtanx/x-limsin/x=0
Your calculation is correct

What is LIM (x → 0) (x-tanx) / (x-sinx)

(x-tanx) / (x-sinx) = (x-sinx / cosx) / (x-sinx) = (xcosx SiNx) / ((x-sinx) cosx) the simplest way is to use series expansion to do SiNx = x-x ^ 3 / 6 + O (x ~ 3) cosx = 1-x ~ 2 / 2 + O (x ~ 2), then xcosx SiNx = x-x ~ 3 / 2-x + X ~ 3 / 6 + O (x ~ 3)

The limit of LIM (1 + 1 / x) x + 2 (x infinity)

The original formula = LIM (1 + 1 / x) ^ x * LIM (1 + 1 / x) ^ 2 because LIM (1 + 1 / x) ^ x = e, the original formula = e * LIM (1 + 1 / x) ^ 2
And lim (1 + 1 / x) ^ 2 = 1 has the original formula = E