lim 1-cos4x/2sin^2x+xtan^2x x趋近于0 求极限

setqwwwe2022-10-04 11:39:542条回答

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yutao88 共回答了19个问题 | 采纳率94.7%
因为1-cos4x/2sin^2x+xtan^2x
=1-(1-2sin^2x)/2sin^2x+sin^2x/cos^2x
=2sin^2x/2sin^2x+sin^2x/cos^2x
=2/1+1/cos^2x
所以原式=2/2=1
1年前
brunoking 共回答了265个问题 | 采纳率
lim 1-cos4x/2sin^2x+xtan^2x x趋近于0
=lim (1-cos4x)/(2sin^2x+xtan^2x) x趋近于0
=lim 2sin^2x/(2sin^2x+xtan^2x )x趋近于0
=lim 2/(2+x/cos^2x )x趋近于0
=1
1年前

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求lim(x-0)1-cos4x/2sin^2x+xtan^2x的极限
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cxodn123456 共回答了16个问题 | 采纳率100%
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