LIM (x / 4) 1 + sin2x / 1-cos4x=

LIM (x / 4) 1 + sin2x / 1-cos4x=

When x → π / 4, we substitute it directly, and we get
sin2x→1,cos4x→-1
Then the limit is 1
I think the topic might be
LIM (x / 4) 1-sin2x / 1 + cos4x
=lim 1-cos(π/2-2x)/1-cos(π-4x)
=lim (1/2)(π/2-2x)² / (1/2)(π-4x)²
=lim (1/4)(π-4x)² / (π-4x)²
=1/4

LIM (sin2x) / 3x = x tends to infinity

lim(sin2x)/3x
=0 x tends to infinity
Because | sin2x|

How to solve the problem Lim ln (1-sinx) / sin2x x → 0,

Because the numerator denominator is 0 when x = 0, we directly use the Roberta rule to get Lim ln (1-sinx) / sin2x = Lim [(- cosx) / 2 (1-sinx) cos2x] and substitute 0 into Lim ln (1-sinx) / sin2x x → 0 = (- 1) / 2 * 1 * 1 = - 1 / 2