高数使用洛必达法则求此极限，谢谢 高数，求极限，如果不用洛必达法则，或者用的话该怎么用，谢谢，...
\u9ad8\u6570\uff0c\u6d1b\u5fc5\u8fbe\u6cd5\u5219\uff0c\u5982\u56fe\uff0c\u6c42\u6781\u9650\uff0c\u8c22\u8c22\uff01(e^x-e^{-x}-2x)/x^3
=(e^x+e^{-x}-2)/3x^2
=(e^x-e^{-x})/6x
=(e^x+e^{-x})/6
=(1+1)/6
=1/3
\u4ee4t=x+1
\u539f\u5f0f=lim(t->0+) [\u221a\u03c0-\u221aarccos(t-1)]/\u221at
\u5206\u5b50\u6709\u7406\u5316
=lim(t->0+) [\u03c0-arccos(t-1)]/[\u221at*(\u221a\u03c0+\u221aarccos(t-1))]
=lim(t->0+) arccos(1-t)/[\u221at*(\u221a\u03c0+\u221aarccos(t-1))]
=lim(t->0+) arcsin[\u221a(2t-t^2)]/[\u221at*(\u221a\u03c0+\u221aarccos(t-1))]
\u7b49\u4ef7\u65e0\u7a77\u5c0f\u4ee3\u6362
=lim(t->0+) \u221a(2t-t^2)/[\u221at*(\u221a\u03c0+\u221aarccos(t-1))]
=lim(t->0+) \u221a(2-t)/[\u221a\u03c0+\u221aarccos(t-1)]
=\u221a2/2\u221a\u03c0
=1/\u221a(2\u03c0)
先上答案
用洛必达来做这个题目没什么问题，注意计算就行，结果是1/4，并非另一个答案写的1/2。
供参考
ln(1+x)等价于x
sinx/2x
1/2
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绛旓細濡傚浘
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绛旓細绛変环鏃犵┓灏忎唬鎹㈢粨鍚娲涘繀杈炬硶鍒鍙互姹傚嚭缁撴灉銆
绛旓細鍘熷紡鍙栧鏁板悗鏈夛紝鍏舵寚鏁颁负 nln(a^1/n+b^1/n)/2 浠=1/n,褰搉瓒嬩簬鏃犵┓澶 锛宼瓒嬩簬0 鐢辨鏈 lim[ln(a^t/2+b^t)/2]/t锛堝簲鐢缃楀繀濉旀硶鍒锛=lim[锛坙na*a^t+lnb*b^t)/2]/[(a^t/2+b^t)/2]=(lna+lnb)/2=ln鈭氾紙ab锛夌敱姝ゅ師寮=e^ln鈭氾紙ab锛=鈭歛b ...
绛旓細鍘熷紡=lim[(lnx)²]'/x'=lim2(lnx)/x =2*lim(lnx)/x =2*lim(lnx)'/x'=2*lim(1/x)=2*0 =0
绛旓細鍏堟敼鎴愬垎寮忕殑褰㈠紡锛屽啀瀵瑰垎瀛愬拰鍒嗘瘝鍒嗗埆姹傚鍚庯紝杩涗竴姝ユ眰鏋佸笺傛湰棰樺氨瑕佹敼鎴愶紙x-蟺)/cotx锛屽垎瀛愬垎姣嶅悓鏃舵眰瀵煎悗灏辨槸1/锛-cscx*cscx)=-(sinx)^2
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