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Taking x as the base infinitesimal, finding the principal part of (x + SiNx ^ 2) ^ 3 Is [x + sin (x ^ 2)] ^ 3 There is another: sin (x + π / 6) - 1 / 2 is still principal
The one upstairs is basically right. It's just a little skipping
(x+sinx^2)^3
(x+[x+O(x^2)]^2)^3
(x+x^2+O(x^3))^3
(x+O(x^2))^3
X ^ 3 + O (x ^ 4), the main part is x ^ 3
the second
sin(x+π/6)-1/2
=sinx cos(π/6)+cosx sin(π/6)-1/2
=Radical 3 / 2 * SiNx + 1 / 2 * cosx - 1 / 2
Radical 3 / 2 * (x + O (x ^ 3)) + 1 / 2 * (1-O (x ^ 2)) - 1 / 2
Radical 3 / 2 * x + 1 / 2-1 / 2 + O (x ^ 2)
Root 3 / 2 * x + O (x ^ 2), the main part is root 3 / 2 * X
Let f (x) be continuous at x = 0, and limx - > 0f (x) - 1 / x = a (a is a constant), find f (0), f '(0)
Obviously, for the limit limx - > 0 [f (x) - 1] / x,
When x tends to 0, its denominator x tends to 0
So if there is a limit, obviously the molecule has to go to zero,
That is, f (x) - 1 = 0, so f (0) = 0
From the law of lobida, we can know that the limit value is equal to the derivation of the denominator of the molecule at the same time
That is limx - > 0 [f (x) - 1] / x = limx - > 0 f '(x) / 1 = a
So limx - > 0, f '(x) = a, that is, f' (0) = a
If the function value f (0) = 0 and the limit limx tends to 0f (x / 2) / x = 2, then the derivative value f '(0)
If the function value f (0) = 0 and the limit limx tends to 0, f (x / 2) / x = 2, then the derivative value f '(0)
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RELATED INFORMATIONS
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- 9. Given LIM (x - > 0) (2arctanx ln (1 + X / 1-x)) / x ^ n = C! = 0, find the values of constants C and N. the solution of this problem is done by using the law of Robida, My method is to split (2arctanx ln (1 + X / 1-x)) / x ^ n into (2arctanx / x ^ n) - (LN (1 + X / 1-x)) / x ^ n, and then use the equivalent infinitesimal to solve it. Why is the result different from the original solution? When is the appropriate time to use the equivalent infinitesimal
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- 11. When x tends to zero, which is higher order infinitesimal compared with SiNx (TaNx + x ^ 2)
- 12. Try to determine the values of constants a, B and C such that e ^ x * (1 + BX + CX ^ 2) = 1 + ax + ο (x ^ 3), where ο (x ^ 3) is the infinitesimal of higher order than x ^ 3 when x → 0 There is an answer (e^x)*(1+Bx+Cx^2)=(1+x+x^2/2!+x^3/3!+o(x^3))*(1+Bx+Cx^2)=1+(1+B)x+(1/2+B+C)x^2+(1/6+B/2+C)x^3+o(x^3)=1+Ax+ο(x^3) So, there are 1 + B = a, 1 / 2 + B + C = 0, 1 / 6 + B / 2 + C = 0 The solution is: a = 1 / 3, B = - 2 / 3, C = 1 / 6 But I can't understand the second equal sign
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- 14. Let's ask a limit problem. If a (x) is an infinitesimal of higher order of X, then when x approaches zero, a (x) / x = 0, and a (x) * x / x = 0, right?
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- 16. Find the maximum and minimum values of the function f (x) = [(SiNx) ^ 4 + (cosx) ^ 4 + (sinxcosx) ^ 2] / (2-2sin2x), and point out the value of X when the maximum value is taken?
- 17. How to simplify SiNx ^ 4 + cosx ^ 4, What about SiNx ^ 4-cosx ^ 4
- 18. Why is the x power of E plus the negative x power of e equal to the 2x power of E plus one
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