![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
The limit of cosx when x tends to π / 2 By the way, we also find the limit of cosx when x tends to zero I just learned the limit, I don't know anything. It should be proved by the limit's ε - δ formula
ε - δ language
For any ε > 0, take δ = ε, when 0
RELATED INFORMATIONS
- 1. X tends to π, (1 + cosx) / (x - π) ^ 2, find the limit
- 2. What is the limit of cosx when x tends to infinity
- 3. What is the limit of cosx when x tends to zero? Ask for detailed explanation
- 4. What is the limit of cosx / X when x tends to zero?
- 5. The limit of cosx / (1-sinx) ^ 2 when x tends to π / 2
- 6. Given the function f (x) = MX ^ 3 + NX ^ 2 (m n belongs to the real number m > N and is not equal to 0), the tangent of the image at (2, f (2)) is parallel to the X axis Finding the relation between M and n
- 7. Lim ln (1 + TaNx) / X when x tends to zero The existence criterion of limit and two important limits are used to solve the problem
- 8. lim(x→+π/2) ln(x-π/2)/tanx
- 9. The x power of E is equal to one part of X to find X
- 10. If the x power of E is equal to x, what is x equal to
- 11. Think about the problem, to know cosx + SiNx = 1, find the nth power of (cosx) + (SiNx)=
- 12. Let f (x) = (SiNx ^ 2 + 1), find the Taylor formula with Peano remainder of F (x) at x = 0, and find f (n) (0)
- 13. When x approaches zero, (1-cosx / 2) is a higher order infinitesimal of X?
- 14. When x → 0, (1-ax ^ 2) ^ (1 / 4) - 1 and xsinx are equivalent infinitesimals
- 15. If x tends to 0, (1-ax ^ 2) ^ 1 / 4 and xsinx are equivalent infinitesimals, find a
- 16. Let x → 0, etanx ex and xn be infinitesimals of the same order, then n is () A. 1B. 2C. 3D. 4
- 17. 16. When x → 0, 2x + (x ^ 2) sin1 / X is the? A equivalent infinitesimal of X, B the same order but not equivalent infinitesimal, C the higher order infinitesimal, d the lower order infinitesimal
- 18. Prove with the definition of function limit: limx ^ 3 (x tends to 2) = 8
- 19. It is proved that the limit x ^ 2 + 2x + 4 = 12 when x tends to 2 Proof by the definition of the limit of higher number function My proof process is that there exists ε > 0, so that | x ^ 2 + 2x + 4-12 | = | X-2 | x + 4|
- 20. How to prove that the limit of x ^ 2 is 4 when x tends to 2 Especially that idea, how did you think of taking δ as that value