题目

$$
\text{求极限 }
\lim_{x\to0}\frac{x\sin x^2 - 2(1 - \cos x)\sin x}{x^5}
$$

解答

$$
x\sin x^2 = x^3 - \dfrac{1}{6}x^6 + o(x^6) \quad
$$

$$
2(1 - \cos x) \sin x = (x^2 - \dfrac{1}{12}x^4 + o(x^4)) \cdot (x - \dfrac{1}{6}x^3 + o(x^3)) = x^3 - (\frac{1}{6} + \frac{1}{12}) x^5 + o(x^5)
$$

$$
\lim_{x\to0}\frac{x\sin x^2 - 2(1 - \cos x)\sin x}{x^5} =
\lim_{x\to0}\frac{x^3 - \dfrac{1}{6}x^6 - x^3 + \dfrac{1}{4}x^5 + o(x^5)}{x^5} = \frac{1}{4}
$$