â´x=1æ¯å¯å»é´æç¹ï¼åªéå®ä¹y(1)=1/3,该å½æ°å¨x=1å¤å°±è¿ç»äºã
å³å½a=-2æ¶f(x)å¨x=0å¤è¿ç»ã
因为:x-1=(x^(1/3)-1)(x^(2/3)+x^(1/3)+1)
带入得
原式=lim(x^(1/3)-1)/[(x^(1/3)-1)(x^(2/3)+x^(1/3)+1)]
=lim1/(x^(2/3)+x^(1/3)+1)
=1/3
所以x=1属于可去间断点
2. 在x=0点连续,则
limf(x=0+)=limf(x=0-)=limf(x=0)
则lim xsin(1/x)=a+2
而x→0时, x为无穷小量,|sin(1/x)|<=1,属于有界函数。由定理可知
无穷小乘以有界函数极限为0.
所以 lim xsin(1/x)=0,则
a+2=0
a=-2