The numerator is 1 - (x) 2 and the denominator is 1 + X + (x) 2. Find the derivative result

The numerator is 1 - (x) 2 and the denominator is 1 + X + (x) 2. Find the derivative result

f(x)=(1-x^2)/(1+x+x^2) df/dx=((1+x+x^2)d(1-x^2)-(1-x^2)d(1+x+x^2))/(1+x+x^2)^2=(-1-4x-x^2)dx/(1+2x+3x^2+2x^3+x^4)

Derivation Lim = (e ^ x-e ^-x) ^ 2 numerator x-0 ln (1 + x ^ 2) denominator

=lim e^(-2x)·(e^(2x) -1) ² / ln(1+x^2)
=lim e^(-2x)· lim(e^(2x) -1) ² / ln(1+x^2)
=1 × lim(2x) ² / Ln (1 + x ^ 2) [Equivalent Infinitesimal Substitution: when x → 0, e ^ X - 1 x]
=lim(2x) ² / (x ^ 2) [Equivalent Infinitesimal Substitution: ln (1 + x) x when x → 0]
=4

Lim x-4, the denominator is X-2 under the root minus 2 under the root, and the numerator is 2x + 1 under the root minus 3. There is no need to calculate the derivative method, because I haven't learned it

[√ (2x + 1) - 3] / [√ (X-2) -√ 2] (the numerator and denominator are multiplied by their rational factors for rational operation)
=[√(2x+1)-3][√(x-2)+√2][√(2x+1)+3]/{[√(x-2)-√2][√(2x+1)+3][√(x-2)+√2]}
=[√(x-2)+√2](2x-8)/{[√(2x+1)+3](x-4)}
=2[√(x-2)+√2]/[√(2x+1)+3]
Therefore, the original formula limit = LIM (x → 4) 2 [√ (X-2) + √ 2] / [√ (2x + 1) + 3] = 2 * 2 √ 2 / (3 + 3) = 2 √ 2 / 3

Find LIM (x →∞) [(√ 2 - √ (1 + cos x)) / (√ (1 + X ²）－ 1) ]. Answer: 1 / (2 √ 2). Seek the process, thank you!

X tends to infinity, right? It should be that x tends to 0. When x tends to 0, the numerator denominator tends to 0. Use lobida's law to derive the numerator denominator at the same time, and get LIM (x → 0) [√ 2 - √ (1 + cos x)] / [√ (1 + x) ²) - 1]=lim(x→0) -1/[2√(1+cos x)] *(-sinx) / [x/√(1+x
...

lim（x→1）（x-1）/（cosπx/2） Can this problem be solved by lobida's law? If so, how should it be calculated?

Can use
lim(x→1)(x-1)/cos(πx/2)
=lim(x→1)1/[-sin(πx/2)*(π/2)]
=1/[-sin(π/2)*(π/2)]
=-2/π

lim（x→1）[（cos（π/2）x]/（1-x）

lim（x→1）[（cos（π/2）x]/（1-x）
=
lim（x→1）[（-π/2sin（π/2）x]/（-1）
=π/2