LIM (x tends to 0) f (x) / x = 2, then LIM (x tends to 0) sin2x / F (3x) =?

LIM (x tends to 0) f (x) / x = 2, then LIM (x tends to 0) sin2x / F (3x) =?

LIM (x tends to 0) sin2x / F (3x) = Lim sin2x * 3x / (f (3x) * 3x) = LIM (3x / F (3x)) * (sin2x / 3x) = Lim 1 / 2 * 2 / 3 = 1 / 3

lim (1-x^2-e^(-x^2))/(sin2x)^4
The problem is (1 minus the square of x minus the square of e minus the square of x) divided by the fourth power of sin2x

LIM (1-x ^ 2-e ^ (- x ^ 2)) / (sin2x) ^ 4 = LIM (1-x ^ 2-e ^ (- x ^ 2)) / (2x) ^ 4 (sin2x ~ 2x, Infinitesimal Substitution) = LIM (- 2x + 2x * e ^ (- x ^ 2)) / (16x ^ 3) (derivative of lobita) = Lim [e ^ (- x ^ 2) - 1] / (8x ^ 2) (reduction) = lim - 2x * e ^ (- x ^ 2) / (16x) (derivative of lobita

Finding the limit LIM (x → π / 4) 1 + sin2x / 1-cos4x

When x → π / 4, we substitute it directly to get sin2x → 1, and cos4x → - 1, then the limit is 1. I think the title may be LIM (x / 4) 1-sin2x / 1 + cos4x = Lim 1-cos (π / 2-2x) / 1-cos (π - 4x) = LIM (1 / 2) (π / 2-2x) & # 178; / (1 / 2) (π - 4x) & # 178; = LIM (1 / 4) (π - 4x) & # 178; / (π -