Proof: the fourth power of SiNx + the fourth power of cosx is equal to the square of 1-2 SiNx multiplied by the square of cosx

Proof: the fourth power of SiNx + the fourth power of cosx is equal to the square of 1-2 SiNx multiplied by the square of cosx

The fourth power of SiNx + the fourth power of cosx = (the square of SiNx + the square of cosx) - the square of 2 SiNx multiplied by the square of cosx = the square of 1-2 SiNx multiplied by the square of cosx

It is proved that 1-2sinxcosx / Cos2 power x-sin2 power x = 1-tanx / 1 + TaNx

(1-2sinxcosx) / (COS? X-sin? X) = (COS? X + sin? X-2sinxcosx) / (COS? X-sin? X) = (cosx SiNx) 2 / (cosx SiNx) (cosx + SiNx) = (cosx SiNx) / (cosx + SiNx) (numerator and denominator divided by cosx) = (cosx / cosx SiNx /)

If the SiN4 power x-cos4 power x = - four fifths, then sin 2x= Please don't jump,

SiN4 power x-cos4 power x = - four fifths
sin^4x-cos^4x=-4/5
(sin^2x+cos^2x)(sin^2x-cos^2x)=-4/5
1*(sin^2x-cos^2x)=-4/5
sin^2x-cos^2x=-4/5
cos^2x-sin^2x=4/5
Cosine double angle formula
∴sin2x=±3/5

By using the property of equivalent infinitesimal, the m power of limit LIM (x tends to 0) sin (the nth power of x) / (SiNx) is obtained (n, m are positive integers)

SiNx is equivalent to X,
Sin (x ^ n) is equivalent to x ^ n
The substitution is approximately x ^ (n-m)
When n = m, 1
When n > m, 0
When n
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Find the limit LIM (x tends to 0) TaNx SiNx / X3 power,

LIM (x tends to 0) TaNx SiNx / X3 power
=lim(x->0)tanx(1-cosx)/x³
=lim(x->0)(x·x²/2)/x³
=1/2

Lim x → 0 SiNx TaNx SiNx Thank you for helping me write more than 3 books

TaNx SiNx = SiNx (1 / cosx-1) (TaNx SiNx) / (SiNx) ^ 3 = (1 / cosx-1) / (SiNx) ^ 2 using the law of l'urbida, the derivation of the denominator = (1 / cosx-1) '/ [(SiNx) ^ 2]' = (SiNx / (cosx) ^ 2) / (2sinxcosx) = 1 / [2 (cosx) ^ 3] = 1 / 2 (cos0 = 1)

Using the property of equivalent infinitesimal to calculate the limit of LIM (x tends to 0) TaNx SiNx / sin cube x

When x tends to 0, TaNx SiNx is equivalent to (x ^ 3) / 2, SiNx is equivalent to x, (SiNx) ^ 3 is equivalent to x ^ 3, so the limit is 1 / 2

Find the limit LIM (X -- > 0) (TaNx SiNx) / [(sin ^ 3) x]

(tanx-sinx)/sin³x
=(sinx/cosx-sinx)/sin³x
=(1/cosx-1)/sin²x
=[(1-cosx)/cosx]/(1-cos²x)
=1/[cosx(1+cosx)]
So the limit = 1 / [1 * (1 + 1)] = 1 / 2

The third power of limit problem Lim x → 0 (TaNx SiNx) / X

Taylor expansion is used
sinx=x-x^3/6+o(x^5)……
tanx=x+x^3/3+o(x^5)……
lim x→0(tanx-sinx)/x^3=lim x→0(x+x^3/3-x+x^3/6+o(x^5))/x^3=lim x→0(x^3/2+o(x^5))/x^3=1/2

Calculate LIM (X -- > 0) [(TaNx SiNx) / sin (x ^ 3)] There are two algorithms Algorithm 1: when X -- > 0, TaNx and X are equivalent infinitesimals, SiNx and X are equivalent infinitesimals, and sin (x ^ 3) and x ^ 3 are equivalent infinitesimals Then the original formula = LIM (X -- > 0) [TaNx / sin (x ^ 3)] - LIM (X -- > 0) [SiNx / sin (x ^ 3)] =lim(x-->0)(x/x^3)-lim(x-->0)(x/x^3)=0 Algorithm 2: when X -- > 0, SiNx and X are equivalent infinitesimals, 1-cosx and (x ^ 2) / 2 are equivalent infinitesimals, and sin (x ^ 3) and x ^ 3 are equivalent infinitesimals Then the original formula = LIM (X -- > 0) [(SiNx / cosx SiNx) / sin (x ^ 3)] =lim(x-->0)(sinx/cosx)(1-cosx)/sin(x^3)] =lim(x-->0)[(x/cosx)(x^2/2)/x^3] =lim(x-->0)[(x^3/2cosx)/x^3] =lim(x-->0)[1/2cosx]=1/2 The answer is that the second algorithm is correct, so what's wrong with the first algorithm? That is to say, even if two identical limit expressions tend to be positive infinity, then their subtraction is meaningless, that is, it can not be said that = 0.

The first algorithm is obviously wrong. LIM (X -- > 0) (x / x ^ 3) = LIM (X -- > 0) [1 / (x ^ 2)] = ∞ limit does not exist. The second algorithm is indeed right, and it is also a general method to solve this kind of problem