lim(x→0)(1-cosx)/(xsinx)=?

lim(x→0)(1-cosx)/(xsinx)=?

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这是0/0型，可以用洛必达法则
lim(x→悄圆0)(1-cosx)/(xsinx)
=lim(x→0)sinx/(sinx+xcosx)
=lim(x→0)1/(1+x/tanx)
只需求出lim(x→0)x/tanx即可
这也是0/0型，也可以用洛必达法则模丛
lim(x→0)x/tanx
=lim(x→0)1/(secx)^2
=1
所以原极限=1/旦运樱2

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1/2

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lim(x→0)(1-cosx)/蠢做掘(xsinx)
=lim(x→带核0)(1-(1-2(sin x/2)^2)/(xsinx)
=(1-(1-2*x^2*(1/2)^2))/x^2
=1/胡差2

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1-cosx(x->0)等价于x^2/2
sinx/x(x->0)=1
所以1/2

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