$f(x) = \cos x$ とおくとき、極限 $\lim_{x \to 0} \frac{1 - f(x)f(10x)f(100x)}{x^2}$ を求める。解析学極限Taylor展開三角関数2025/4/201. 問題の内容f(x)=cosxf(x) = \cos xf(x)=cosx とおくとき、極限 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) を求める。2. 解き方の手順cosx\cos xcosx の Taylor展開を考える。cosx=1−x22+O(x4)\cos x = 1 - \frac{x^2}{2} + O(x^4)cosx=1−2x2+O(x4) である。f(x)=cosxf(x) = \cos xf(x)=cosx を与えられた極限に代入する。limx→01−cosxcos(10x)cos(100x)x2\lim_{x \to 0} \frac{1 - \cos x \cos(10x) \cos(100x)}{x^2}limx→0x21−cosxcos(10x)cos(100x)cosxcos(10x)cos(100x)≈(1−x22)(1−(10x)22)(1−(100x)22)=(1−x22)(1−50x2)(1−5000x2)\cos x \cos(10x) \cos(100x) \approx (1 - \frac{x^2}{2})(1 - \frac{(10x)^2}{2})(1 - \frac{(100x)^2}{2}) = (1 - \frac{x^2}{2})(1 - 50x^2)(1 - 5000x^2)cosxcos(10x)cos(100x)≈(1−2x2)(1−2(10x)2)(1−2(100x)2)=(1−2x2)(1−50x2)(1−5000x2)≈1−x22−50x2−5000x2=1−(12+50+5000)x2=1−1+100+100002x2=1−101012x2\approx 1 - \frac{x^2}{2} - 50x^2 - 5000x^2 = 1 - (\frac{1}{2} + 50 + 5000)x^2 = 1 - \frac{1 + 100 + 10000}{2}x^2 = 1 - \frac{10101}{2}x^2≈1−2x2−50x2−5000x2=1−(21+50+5000)x2=1−21+100+10000x2=1−210101x2limx→01−(1−101012x2)x2=limx→0101012x2x2=101012\lim_{x \to 0} \frac{1 - (1 - \frac{10101}{2}x^2)}{x^2} = \lim_{x \to 0} \frac{\frac{10101}{2}x^2}{x^2} = \frac{10101}{2}limx→0x21−(1−210101x2)=limx→0x2210101x2=2101013. 最終的な答え101012\frac{10101}{2}210101