当x→0时下列无穷小是x的n阶无穷小,求阶数n: (I) (Ⅱ)(1+tan2x)sinx一1; (Ⅲ) (Ⅳ)∫0xsint.sin(1一cost)2dt.

admin2017-07-28  30

问题 当x→0时下列无穷小是x的n阶无穷小,求阶数n:
(I)
(Ⅱ)(1+tan2x)sinx一1;
(Ⅲ)
(Ⅳ)∫0xsint.sin(1一cost)2dt.

选项

答案(I)[*]一1~x4—2x2~一2x2 (x→0),即当x→0时[*]—1是x的2阶无穷小,故n=2. (Ⅱ)(1+tan2x)sinx一1~ln[(1+tan2x)sinx一1+1] =sinxln(1+tan2x)~sinxtan2x~x.x2=x3 (x→0),即当x→0时(1+tan2x)sinx一1是x的3阶无穷小,故n=3. (Ⅲ)由[*]是x的4阶无穷小,即当x→0时[*]是x的4阶无穷小,故n=4. [*] 即当x→0时∫0xsintsin(1一cost)2dt是x的6阶无穷小,故n=6.

解析
转载请注明原文地址:https://jikaoti.com/ti/5YwRFFFM
0

最新回复(0)