求极限lim x→π/2 (1-sinx)/(cos^2)x如图.请说明计算时使用的公式名称.
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![求极限lim x→π/2 (1-sinx)/(cos^2)x如图.请说明计算时使用的公式名称.](/uploads/image/z/1967123-11-3.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90lim+x%E2%86%92%CF%80%2F2+%281-sinx%29%2F%28cos%5E2%29x%E5%A6%82%E5%9B%BE.%E8%AF%B7%E8%AF%B4%E6%98%8E%E8%AE%A1%E7%AE%97%E6%97%B6%E4%BD%BF%E7%94%A8%E7%9A%84%E5%85%AC%E5%BC%8F%E5%90%8D%E7%A7%B0.)
求极限lim x→π/2 (1-sinx)/(cos^2)x如图.请说明计算时使用的公式名称.
求极限lim x→π/2 (1-sinx)/(cos^2)x
如图.
请说明计算时使用的公式名称.
求极限lim x→π/2 (1-sinx)/(cos^2)x如图.请说明计算时使用的公式名称.
lim(x→π/2)(1-sinx)/(cosx)^2
=lim(x→π/2)(1-sinx)/[1-(sinx)^2]
=lim(x→π/2)(1-sinx)/[(1-sinx)(1+sinx)]
=lim(x→π/2)[1/(1+sinx)]
=1/2
没用什么公式,有不懂的,再补充吧……
原式=LIM(1-SINX)/[1-(SIN^2)X]
=LIM(1-SINX)/(1+SINX)(1-SINX)
=LIM 1/(1+SINX)
=1/2
解法一:原式=lim(x->π/2)[-cosx/(-2sinxcosx)] (0/0型,应用罗比达法则)
=lim(x->π/2)[1/(2sinx)]
=1/(2*1)
=1/2
解法二:原式=lim(x->π/2)[(1-sinx)/(1-sin²x)]
...
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解法一:原式=lim(x->π/2)[-cosx/(-2sinxcosx)] (0/0型,应用罗比达法则)
=lim(x->π/2)[1/(2sinx)]
=1/(2*1)
=1/2
解法二:原式=lim(x->π/2)[(1-sinx)/(1-sin²x)]
=lim(x->π/2){(1-sinx)/[(1+sinx)(1-sinx)]}
=lim(x->π/2)[1/(1+sinx)]
=1/(1+1)
=1/2
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