<s>[INST] Find constants $a$ and $b$ such that $ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$ 
Find constants $a$ and $b$ such that $$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$$

First,$a$ should be positive to make sure the limit is meaningful as $x \to 0^-$ .
Then I check the limit of the numerator,say $- \frac{a}{2}<x<0$. For t in the interval(x,0), there's  $\frac{1}{\sqrt{a+t}}<M$, where $M>0$,so $$| \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} | <M |\int^x_0t^2dt| =M \frac{-x^3}{3},$$ by using the sandwich theorem I get $\lim_{x \to 0^-} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} =0$,thing should be the same when $x>0$ .
So the limit is the form $0/0$ . Apply L'Hopital's rule twice I get 
$$\begin{align} \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}  &= \lim_{x \to 0} \frac{x^2}{ \sqrt{a+x}(bx-\sin x)} \\ &= \lim_{x \to 0} \frac{2x} { \frac{bx-\sin x}{2 \sqrt{a+x}}+ \sqrt{a+x}  (b- \cos x)} \\ &= \lim_{x \to 0} \frac{4x \sqrt{a+x}}{3bx+2ab-2(a+x)\cos x -\sin x} \end{align}$$Now the limit becomes $ \frac{0}{2ab-2a}$,so $2a(b-1)=0$ and $b=1$ cause $a$ is positive.
Apply L'Hopital's rule again $$\begin{align}  \lim_{x \to 0} \frac{4x \sqrt{a+x}}{3x+2a-2(a+x)\cos x -\sin x} &= \lim_{x \to 0} \frac{4( \sqrt{a+x} + \frac{x}{2 \sqrt{a+x}})}{3+2a \sin x- 2 \cos x  + 2 x \sin x - \cos x} \\ & =  \lim_{x \to 0} \frac{2(2a+3x)}{\sqrt{a+x} (3-3 \cos x  + 2a \sin x + 2x \sin x )}  \\ &= \frac{4a}{\sqrt{a}  0}  \quad ?!\end{align}$$
and I fail to solve it.
Thanks in advance.
 [/INST] Let's substitute $t\mapsto tx$:
$$
\begin{align}
\lim_{x\to0}\frac1{bx-\sin(x)}\int_0^x\frac{t^2\,\mathrm{d}t}{\sqrt{a+t}}
&=\lim_{x\to0}\frac{x^3}{bx-\sin(x)}\color{#00A000}{\int_0^1\frac{t^2\,\mathrm{d}t}{\sqrt{a+tx}}}\\
&=\color{#00A000}{\frac1{3\sqrt{a}}}\lim_{x\to0}\frac{x^3}{bx-\sin(x)}
\end{align}
$$
If $b\ne1$, the limit on right would be $0$. Thus, we need $b=1$.
In this answer, it is shown that $\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16$, therefore, we want
$$
\frac1{3\sqrt{a}}\cdot6=1\implies a=4
$$
</s>