解ï¼ï¼1ï¼å·¦æé=0ï¼
å³æé=limï¼xâ0+)[â(x+1)-1]/âx(0/0å½¢å¼ï¼ç¨æ´å¿
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=limï¼xâ0+) {1/[2â(x+1)]}/(1/âx)=limï¼xâ0+) âx/[2â(x+1)](代å
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å·¦æé=å³æéï¼å½æ°å¨x=0å¤æå®ä¹ï¼f(0)=0; æ以å½æ°è¿ç»ã
(2)左导æ°=0ï¼
å½x>0æ¶ï¼f'(x)={âx/[2â(x+1)]-[â(x+1)-1]/(2âx)}/x=[x-(x+1)+â(x+1)]/{2xâ[x(x+1)]}
=[â(x+1)-1]/[2xâ(x^2+x)];
å³å¯¼æ°=limï¼xâ0+)f'(x)=limï¼xâ0+)[â(x+1)-1]/[2xâ(x^2+x)]
=limï¼xâ0+){1/[2â(x+1)]/[2â(x^2+x)+x*(2x+1)/â(x^2+x)]
=limï¼xâ0+) 1/{4â[(x^2+x)(x+1)]+2x*(2x+1)}=+âï¼
左导æ°â å³å¯¼æ°ï¼å¯¼æ°ä¸åå¨ã
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å³å¯¼æ°=lim(xâ0+)[f(x)-f(0)]/x=lim(xâ0+){ [â(x+1)-1]/âx-0}/x
=lim(xâ0+) [â(x+1)-1]/(xâx)(ååç¨éº¦å
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=lim(xâ0+) {[(1+x/2+(1/2)(1/2-1)x^2/2!+o(x^3)]-1}/âx^3
=lim(xâ0+) [x/2-x^2/8+o(x^3)]/âx^3(ç¥å»é«é¶æ ç©·å°)=lim(xâ0+) 1/(2âx)=+âï¼
左导æ°â å³å¯¼æ°ï¼å¯¼æ°ä¸åå¨ãåé¢è¿ç§æ¹æ³ä¸è¬äººè®°ä¸ä½éº¦å
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