Let $a$ and $b$ denote two positive constants. If
$$\lim_{x\to 0} \Bigg(\frac{\int_0^x \frac{t^2}{\sqrt{a+2t^5}}dt}{bx-e\sin x}\Bigg)= \frac{1}{\pi},$$
find the value of $a$.
What I've tried:
I've used L'Hospital's rule since it is a $\frac00$ form, to get
$$\frac{\lim_{x\to 0} \Big(\frac{x^2}{\sqrt{a+2x^5}}\Big)}{{\lim_{x\to 0} ( b-e\cos x)}}$$
If $x$ tends to $0$, the equation will become $\frac{\frac{0}{\sqrt{a}}}{b-e}$, making it impossible to become $\frac{1}{\pi}$.
Did I make a mistake somewhere?
Set $b = e$ then you still have the form $0/0$. Apply L'Hospital 2 more times you will get \begin{equation} \lim_{x \rightarrow 0} \frac{\int_0^x\frac{t^2dt}{\sqrt{a + 2t^5}}}{b - e \sin(x)} = \frac{2}{e\sqrt{a}}. \end{equation} Therefore $a = 4\pi^2/e^2$. Note that it is important that the final answer is $a \neq 0$ otherwise we have to start again without the assumption that $\lim_{x\rightarrow 0}\int^x t^2/\sqrt{a - 2t^5}dt = 0$. But in fact if $a = 0$ then the integral $= \int^x_0 dt/\sqrt{2t} = \sqrt{2x}$. So we can in fact see that in this case the limit is $\infty \neq 1/\pi$.