$f(x) = \cos x$ とするとき、$\lim_{x \to 0} \frac{1 - f(x)f(10x)f(100x)}{x^2}$ を求めよ。解析学極限三角関数テイラー展開ロピタルの定理2025/4/201. 問題の内容f(x)=cosxf(x) = \cos xf(x)=cosx とするとき、limx→01−f(x)f(10x)f(100x)x2\lim_{x \to 0} \frac{1 - f(x)f(10x)f(100x)}{x^2}limx→0x21−f(x)f(10x)f(100x) を求めよ。2. 解き方の手順f(x)=cosxf(x) = \cos xf(x)=cosx を代入すると、求める極限はlimx→01−cosxcos(10x)cos(100x)x2 \lim_{x \to 0} \frac{1 - \cos x \cos(10x) \cos(100x)}{x^2} x→0limx21−cosxcos(10x)cos(100x)である。1−cosx=2sin2x21 - \cos x = 2\sin^2 \frac{x}{2}1−cosx=2sin22x を用いると、limx→01−cosxx2=12\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}limx→0x21−cosx=21 が成り立つ。また、cosx=1−x22+O(x4)\cos x = 1 - \frac{x^2}{2} + O(x^4)cosx=1−2x2+O(x4) である。1−cosxcos(10x)cos(100x)=1−(1−x22+O(x4))(1−(10x)22+O(x4))(1−(100x)22+O(x4))1 - \cos x \cos(10x) \cos(100x) = 1 - (1 - \frac{x^2}{2} + O(x^4))(1 - \frac{(10x)^2}{2} + O(x^4))(1 - \frac{(100x)^2}{2} + O(x^4))1−cosxcos(10x)cos(100x)=1−(1−2x2+O(x4))(1−2(10x)2+O(x4))(1−2(100x)2+O(x4))=1−(1−x22−100x22−10000x22+O(x4))= 1 - (1 - \frac{x^2}{2} - \frac{100x^2}{2} - \frac{10000x^2}{2} + O(x^4))=1−(1−2x2−2100x2−210000x2+O(x4))=x22+100x22+10000x22+O(x4)= \frac{x^2}{2} + \frac{100x^2}{2} + \frac{10000x^2}{2} + O(x^4)=2x2+2100x2+210000x2+O(x4)=101012x2+O(x4)= \frac{10101}{2}x^2 + O(x^4)=210101x2+O(x4)したがって、limx→01−cosxcos(10x)cos(100x)x2=limx→0101012x2+O(x4)x2=101012 \lim_{x \to 0} \frac{1 - \cos x \cos(10x) \cos(100x)}{x^2} = \lim_{x \to 0} \frac{\frac{10101}{2}x^2 + O(x^4)}{x^2} = \frac{10101}{2} x→0limx21−cosxcos(10x)cos(100x)=x→0limx2210101x2+O(x4)=2101013. 最終的な答え101012\frac{10101}{2}210101