解ï¼å享ä¸ç§è§£æ³ãâµsin3x=3sinx-4(sinx)^3ï¼cosxcos2xcos3x=(1/2)(cos3x+cosx)cos3x=(1/4)(1+cos6x+cos4x+cos2x)ï¼â´3sinx-sin3x=4(sinx)^3=2sinx(1-cos2x)ï¼1-cosxcos2xcos3x=(1/4)(3-cos6x-cos4x-cos2x)ãæ
ï¼åå¼=(1/8)lim(xâ0)(3-cos6x-cos4x-cos2x)/(1-cos2x)ï¼å±â0/0âåï¼ç¨æ´æ¯å¡æ³åï¼æåå¼=(1/8)lim(xâ0)(6sin6x+4sin4x+2sin2x)/(2sins2x)=(1/8)lim(xâ0)(3sin6x/sins2x+2sin4x/2sins2x+1)ï¼èxâ0æ¶ï¼sin6x~6xï¼sins4x~4xï¼sin2x~2xï¼â´åå¼=(1/8)(3*3+2*2+1)=7/4ããæ¬é¢ï¼å¦è¥ç¨
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