设函数f(x)在点x0处可导,f(x0)=0,f 设函数f(x)在点x0处可导,且f(x0)!=0,求极限li...
\u8bbe\u51fd\u6570f\uff08x\uff09\u5728\u70b9x0\u5904\u53ef\u5bfc\uff0c\u5219lim\u4e28\u25b3x\u21920 f\uff08x0-2\u25b3x\uff09-f\uff08x0\uff09/\u25b3x=\uff1f\u5c06\u53d8\u91cf\u7a0d\u4f5c\u66ff\u6362\uff0c\u8be6\u89c1\u4e0b\u56fe\uff0c\u671b\u91c7\u7eb3\u3002
\u3000\u3000\u5148\u8ba1\u7b97\u53d6\u5bf9\u6570\u540e\u7684\u6781\u9650
\u3000\u3000\u3000lim(n\u2192\u221e)[lnf(x0+1/n)-lnf(x0)]/(1/n)
\u3000\u3000= f'(x0)/f(x0)\uff0c
\u6240\u4ee5
\u3000\u3000\u3000lim(n\u2192\u221e)[f(x0+1/n)/f(x0)]^n
\u3000\u3000= e^lim(n\u2192\u221e)[lnf(x0+1/n)-lnf(x0)]/(1/n)
\u3000\u3000= e^[f'(x0)/f(x0)]\u3002
所以lim(x-->x0+)|f(x)|/x =f`(x0)
lim(x-->x0-)|f(x)|/x =-f`(x0)
因为f`(x0)不等于0,即左右导数不相等,所以不可导
f'(x0)≠0,所以f'(x0)只能是大于或小于0 不管哪种情况 在x0的某个领域内都有 x0两边函数值异号 因此lim(x0+) |f(x)|/x=f(x0) lim(x0-) |f(x)|/x=-f(x0) 左右导数不等 所以在此点不可导
绛旓細=(1/2)f'(x0) + (1/2)f'(x0)= f'(x0)
绛旓細= f'(x0) * (- 1/2)= 2 * (- 1/2)= - 1
绛旓細鍏堣绠楀彇瀵规暟鍚庣殑鏋侀檺銆lim(n鈫掆垶)[lnf(x0+1/n)-lnf(x0)]/(1/n)= f'(x0)/f(x0),鎵浠ャlim(n鈫掆垶)[f(x0+1/n)/f(x0)]^n= e^lim(n鈫掆垶)[lnf(x0+1/n)-lnf(x0)]/(1/n)= e^[f'(x0)/f(x0)]...
绛旓細= f'(x0) * (- 1/2)= 2 * (- 1/2)= - 1
绛旓細璁惧嚱鏁癴锛坸锛夊湪x0澶勫彲瀵锛屽垯 锛1锛夊嚱鏁拌繛缁紱锛2锛夊乏鏋侀檺=鍙虫瀬闄
绛旓細1銆鍑芥暟f锛坸锛夊湪鐐箈0澶勫彲瀵硷紝鐭ュ嚱鏁癴锛坸锛夊湪鐐箈0澶勮繛缁2銆佸嚱鏁癴锛坸锛夊湪鐐箈0澶勫彲瀵硷紝鐭ュ嚱鏁癴锛坸锛夊湪鐐箈0瀛樺湪鍒囩嚎銆3銆佸嚱鏁癴锛坸锛夊湪鐐箈0澶勫彲瀵硷紝鐭ュ嚱鏁癴锛坸锛夊湪鐐箈0澶勬瀬闄愬瓨鍦ㄣ
绛旓細鍘熷紡=lim鈻硏鈫0f(x0?鈻x)?f(x0)?(?鈻硏)=-lim鈻硏鈫0f(x0?鈻硏)?f(x0)?鈻硏=-f鈥诧紙x0锛夛紙2锛塴imh鈫0f(x0+h)?f(x0?h)2h=12limh鈫0 f(x0+h)?f(x0)+f(x0)?f(x0?h)h=12limh鈫0[f(x0+h)?f(x0)h?f(x0?h)?f(x0)?h]=12[f鈥诧紙x0锛+f鈥诧紙x0...
绛旓細鍦ㄧ偣x0澶勫彲瀵硷紝鍒f(x)鍦ㄧ偣x0鐨勬煇閭诲煙鍐呭繀瀹氳繛缁紝杩欏彞璇濇槸閿欒鐨勩備妇渚嬭鏄庯細f(x)=0锛屽綋x鏄湁鐞嗘暟 f(x)=x^2锛屽綋x鏄棤鐞嗘暟 鍙湪x=0澶勭偣杩炵画锛屽苟鍙锛屾寜瀹氫箟鍙獙璇佸湪x=0澶勫鏁颁负0 浣唂(x) 鍦ㄥ埆鐨勭偣閮戒笉杩炵画 鍑芥暟鍙鍒欏嚱鏁拌繛缁紱鍑芥暟杩炵画涓嶄竴瀹氬彲瀵硷紱涓嶈繛缁殑鍑芥暟涓瀹氫笉鍙銆
绛旓細鍏堣绠楀彇瀵规暟鍚庣殑鏋侀檺 lim(n鈫掆垶)[lnf(x0+1/n)-lnf(x0)]/(1/n)= f'(x0)/f(x0)锛鎵浠 lim(n鈫掆垶)[f(x0+1/n)/f(x0)]^n = e^lim(n鈫掆垶)[lnf(x0+1/n)-lnf(x0)]/(1/n)= e^[f'(x0)/f(x0)]銆
绛旓細瑙g瓟锛鍑芥暟y=f(x)鍦ㄧ偣x0澶勫彲瀵硷紝涓攆'(x0)=a 鍒 lim鈻硏鈫0 f(x0鈥撯柍x)鈥揻(x0)/鈻硏 =f'(x0)=a 鈭 lim鈻硏鈫0 f(x0鈥2鈻硏)鈥揻(x0)/2鈻硏 =f'(x0)=a 鈭 lim鈻硏鈫0 f(x0鈥2鈻硏)鈥揻(x0)/鈻硏 =2a ...