当x趋于0时(cosx-1)/x的极限是多少？帮忙证一下，有思路也可以的拜托了各位 谢谢 为什么当x趋向0时，(1-cosx)／x^2的极限是1/2
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(cosx)^(1/x^2)=e^[ln(cosx)^(1/x^2)]
=e^(1/x^2 * lncosx)
=e^(lncosx/x^2)
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\u4e00\u3001\u7b49\u4ef7\u65e0\u7a77\u5c0fln(1+x)\uff5ex,1-cosx\uff5e x^2/2
lim(lncosx/x^2)=lim ln[1+(cosx-1)]/x^2
=lim (cosx-1)/x^2
=lim (-x^2/2)/x^2
=-1/2
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lim(lncosx/x^2)=lim (-sinx/cosx)/2x
=lim (-1/2cosx)
=-1/2
\u6240\u4ee5\u539f\u5f0f=lim e^(lncosx/x^2)
=e^lim(lncosx/x^2)
=e^(-1/2)
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cos2a=1-2(sina)^2
\u22341-cosx=2(sinx/2)^2
\u2234
limx->0 (1-cosx)/x^2
=limx->0 2(sinx/2)^2 /x^2
=limx->0 2(sinx/2)^2 /4*(x/2)^2
=1/2limx->0 (sinx/2)^2 /(x/2)^2
=1/2
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绛旓細娲涘繀杈炬硶鍒欙紝锛坈osx-1锛锛弜²=-sinx/(2x)=-1/2
