求lim(x-0)1-cos4x/2sin^2x+xtan^2x的极限

问题描述:

求lim(x-0)1-cos4x/2sin^2x+xtan^2x的极限

1 - cos4x 等价于 1/2*(4x)^2 = 8x^2
2sin²x + x tan²x
= sin²x ( 2+ x/cos²x )
等价于 x² (2 + x/cos²x)
原式 = 8 / (2+ x/cos²x) = 4