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Last edited by isee 2022-3-14 14:25源自知乎提问
题：求 $$\lim_{x \to 0}\frac {2\int_0^x\sin (t^2x^2)\mathrm dt}{x^5}.$$
因 $0\leqslant t \leqslant x$，当 $x\to 0$ 时， $t\to 0$，于是由泰勒公式 $\sin (x^2t^2)=x^2t^2+o(t^6)$.
于是 $$\int_0^x\sin (t^2x^2)\mathrm dt=\int_0^x x^2t^2+o(t^6)\mathrm dt=\frac 13x^5+o(x^7),$$
则 $$\lim_{x \to 0}\frac {2\int_0^x\sin (t^2x^2)\mathrm dt}{x^5}=\lim_{x \to 0}\frac {2\left(\frac 13x^5+o(x^7)\right)}{x^5}=\frac 23.$$ |
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