f(x)=x-sin(ax)与g(x)=x^2【ln(1-bx)】等价无穷小.求a,b的值.
x→0
人气:344 ℃ 时间:2020-03-22 12:58:02
解答
提到无穷小,必须加上(x→?),这里呢?解考察 L = lim(x→0)[f(x)/g(x)] = lim(x→0){[x-sin(ax)]/[(x^2)ln(1-bx)]}= lim(x→0){[x-sin(ax)]/[(x^2)(-bx)]}= (-1/b)lim(x→0){[x-sin(ax)]/(x^3)}(0/0)= (-1/b)lim(x→0){[1-acos(ax)]/(3x^2)}要最后一个极限存在,需 a =1,因此 L = (-1/3b)lim(x→0){[1-cos(x)]/(x^2)}= (-1/3b)(1/2),令(-1/3b)(1/2) = 1,可解得 b。这是因分母 3x^2 → 0 (x→0),若分子 1-acos(ax)非→ 0 (x→0),则如上极限不存在,因而应有1-acos(ax)→ 0 (x→0),这样需a =1。
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