Computing $\lim\limits_{x\to0}\int_0^x\frac{t^2}{(x-\sin x)\sqrt{a+t}}\,dt$ without L'Hopital I am computing the following limit
$$\lim_{x\to0}\int_0^x\frac{t^2}{(x-\sin x)\sqrt{a+t}}\,dt$$
where $a$ is a parameter. the case $a= 0$ has been resolved. But yet, for the case $ a\neq 0$ I want to  compute it without using L'Hopital rule.
 A: If you know how to calculate the limit of $\frac{x^3}{x-\sin x}$ without L'Hospital, then the rest follows from the mean value theorem for integrals (if we may use it - it is quite similar to L'Hospital/Bernoulli's rule)
$$
\int_0^x\frac{t^2}{x^3\sqrt{a+t}}\,dt=\left[u=\frac{t}{x}\right]=\int_0^1\frac{u^2}{\sqrt{a+xu}}\,du=\frac{1}{\sqrt{a+x\xi}}\int_0^1u^2\,du=\frac{1}{\sqrt{a+x\xi}}\cdot\frac13\to\frac{1}{3\sqrt{a}}.
$$
A: With Maclaurin series:
$$\lim_{x\to0}\int_0^x\frac{t^2}{(x-\sin x)\sqrt{a+t}}\,dt=\lim_{x\to0}\frac{1}{(x-\sin x)}\int_0^x\frac{t^2}{\sqrt{a+t}}\,dt=\lim_{x\to0}\frac{1}{(x-\sin x)}\left[\frac{2}{15}\sqrt{a+t}(8a^2-4at+3t^2)\right]_{0}^{x}=\lim_{x\to0}\frac{1}{(x-\sin x)}\frac{2}{15}\left[\sqrt{a+x}(8a^2-4ax+3x^2)-\sqrt{a}(8a^2)\right]=\frac{2\sqrt{a}}{15}\lim_{x\to0}\frac{\sqrt{1+\frac{x}{a}}(8a^2-4ax+3x^2)-8a^2}{x-\sin x}=\frac{2\sqrt{a}}{15}\lim_{x\to0}\frac{(1+\frac{x}{2a}-\frac{x^2}{8a^2}+\frac{x^3}{16a^3}+O(x^4))(8a^2-4ax+3x^2)-8a^2}{x-x+\frac{x^3}{6}+O(x^4)}=\frac{2\sqrt{a}}{15}\lim_{x\to0}\frac{\frac{3x^3}{2a}+\frac{x^3}{2a}-\frac{5x^3}{16a}+O(x^4)}{\frac{x^3}{6}+O(x^4)}=\frac{2\sqrt{a}}{15}\frac{5}{2a}\frac{1}{\frac{1}{6}}=\frac{2}{\sqrt{a}}$$
