当χ→0时下列无穷小是χ的n阶无穷小,求阶数n: (Ⅰ)-1; (Ⅱ)(1+tan2χ)sinχ-1; (Ⅲ); (Ⅳ)∫0χsint.sin(1-cost)2dt.

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问题 当χ→0时下列无穷小是χ的n阶无穷小,求阶数n:
    (Ⅰ)-1;
    (Ⅱ)(1+tan2χ)sinχ-1;
    (Ⅲ)
    (Ⅳ)∫0χsint.sin(1-cost)2dt.

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答案(Ⅰ)[*]-1~χ4-2χ2~2χ2(χ→0),即当χ→0时[*]-1是χ的2阶无穷小,故n=2. (Ⅱ)(1+tan2χ)sinχ-1~ln[(1+tan2χ)sinχ-1+1] =sinχln(1+tan2χ)-sinχtan2χ~χ.χ2=χ3 (χ→0), 即当χ→0时(1+tan2χ)sinχ-1是χ的3阶无穷小,故n=3. (Ⅲ)由1-[*]是χ的4阶无穷小,即当χ→0时[*]是χ的4阶无穷小,故n=4. (Ⅳ)[*] 即当χ→0时∫0χsintsin(1-cost)2dt是χ的6阶无穷小,故n=6.

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