lim(x趋于派/4)1+sin2x/1-cos4x=

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lim(x趋于派/4)1+sin2x/1-cos4x=lim(x趋于派/4)1+sin2x/1-cos4x=lim(x趋于派/4)1+sin2x/1-cos4x=x→π/4时,直接代入,得sin2x→

lim(x趋于派/4)1+sin2x/1-cos4x=
lim(x趋于派/4)1+sin2x/1-cos4x=

lim(x趋于派/4)1+sin2x/1-cos4x=
x→π/4时,直接代入,得
sin2x→1,cos4x→-1
则极限为1
我想题目可能是
lim(x趋于派/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