当x→0时下列无穷小是x的n阶无穷小,求阶数n: (Ⅰ)ex4-2x2-1; (Ⅱ)(1+tan2x)sinx-1; (Ⅲ) (Ⅳ)∫0xsint.sin(1-cost)2dt.

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问题 当x→0时下列无穷小是x的n阶无穷小,求阶数n:
(Ⅰ)ex4-2x2-1;
(Ⅱ)(1+tan2x)sinx-1;
(Ⅲ)
(Ⅳ)∫0xsint.sin(1-cost)2dt.

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答案(Ⅰ)ex4-2x2-1~x4-2x2~-2x2 (x→0),即 当x→0时ex4-2x2-1是x的2阶无穷小, 故n=2. (Ⅱ)(1+tan2x)sinx-1~ln[(1+tan2x)sinx-1+1] =sinxln(1+tan2x)~sinxtan2x~x.x2=x3 (x→0), 即当x→0时(1+tan2x)sinx-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→时[*]是x的4阶无穷小,故n=4. (Ⅳ)[*] 即当x→0时∫02sintsin(1-cost)2dt是x的6阶无穷小,故n=6.

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