设 $\dps{\lim_{x\to 0}\frac{1}{bx-\sin x}\int_0^x \frac{t^2}{\sqrt{a+t^2}}\rd t=1}$, 试求正常数 $a$ 与 $b$. (华中师范大学)

解答: 由 $$\beex \bea 1&=\lim_{x\to 0}\frac{1}{bx-\sin x}\int_0^x \frac{t^2}{\sqrt{a+t^2}}\rd t\\ &=\lim_{x\to 0}\frac{b-\cos x} \cdot \frac{x^2}{\sqrt{a+x^2}}\\ &=\lim_{x\to 0}\frac{1}{x^2/2} \cdot \frac{x^2}{\sqrt{a}}\quad\sex{b=1} \eea \eeex$$ 知 $a=4$, $b=1$.