I am given the following problem which I have problem to know where to even start:
The question: $\lim_{x\to 0} \frac{\int_0^x\frac{t^2}{\sqrt{a+2t^5}}dt} {bx-esinx}=\frac{1}{\pi}$
The part where I will like to know where to start: $\int_0^x\frac{t^2}{\sqrt{a+2t^5}}dt$
I appreciate suggestions on how I may get about solving the integration part before I move on to solve the limit question as a whole.
On
If $b\neq e$ then dividing the numerator and denominator by $x$ we can see that the limit is $0$ instead of $1/\pi$. Hence we must have $b=e$ and these constants must be non-zero and thus using $(x-\sin x) /x^3\to 1/6$ we see that the given condition implies $$\lim_{x\to 0}\frac{1}{x^3}\int_{0}^{x}\frac{t^2}{\sqrt{a+2t^5}}\,dt=\frac{b}{6\pi}\tag{1}$$ Further note that $a>0$ so that the integral makes sense at $t=0$. If $x>0$ then we have for $0<t<x$ $$\frac{t^2}{\sqrt{a+2x^5}}\leq \frac{t^2}{\sqrt{a+2t^5}}<\frac{t^2}{\sqrt{a}}$$ and integrating this with respect to $t$ on $[0,x]$ we get $$\frac{1}{3\sqrt{a+2x^5}}<\frac{1}{x^3}\int_{0}^{x}\frac{t^2}{\sqrt{a+2t^5}}\,dt<\frac{1}{3\sqrt{a}}$$ and thus by Squeeze Theorem the limit $(1)$ is equal to $3/\sqrt{a}$ and thus from $(1)$ we have $9/a=b^2/36\pi^2$ so $a$ is determined in terms of $b=e$.
You do not need to evaluate the integral.
Simply use L'Hopitals Rule. Integral will vanish on taking derivative.
For taking derivative w.r. t. $x$, use Leibnitz Integral Rule.