Finding Limits 1

The conditions of L'Hopital's Rule are satisfied, so that the required limit is:

$\large\lim_{x\to 0}\large\frac{\frac{d}{dx}\int_0^x( \sin t^3) dt}{\frac{d}{dx}(x^4)}=\lim_{x\to 0}\frac {\sin x^3}{4x^3}= \lim_{x\to 0} \frac{\frac{d}{dx}(\sin x^3)}{\frac{d}{dx}(4x^3)}=\lim_{x\to 0} \frac {3x^2 \cos x^3}{12x^2} = \frac{1}{4}$

$\begin{aligned} L & = \lim_{x \to 0} \frac 1{x^4} \int_0^x \color{#3D99F6}{\sin t^3} \ dt & \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \lim_{x \to 0} \frac 1{x^4} \int_0^x \color{#3D99F6}{\sum_{n=0}^\infty \frac {(-1)^n t^{6n+3}}{(2n+1)!}} \ dt \\ & = \lim_{x \to 0} \frac 1{x^4} \sum_{n=0}^\infty \int_0^x \frac {(-1)^n t^{6n+3}}{(2n+1)!} \ dt \\ & = \lim_{x \to 0} \frac 1{x^4} \sum_{n=0}^\infty \frac {(-1)^n t^{6n+4}}{(6n+4)(2n+1)!} \bigg|_0^x \\ & = \lim_{x \to 0} \frac 1{x^4} \sum_{n=0}^\infty \frac {(-1)^n x^{6n+4}}{(6n+4)(2n+1)!} \\ & = \lim_{x \to 0} \sum_{n=0}^\infty \frac {(-1)^n x^{6n}}{(6n+4)(2n+1)!} \\ & = \frac {(-1)^0}{(6(0)+4)(2(0)+1)!} \\ & = \boxed{\dfrac 14} \end{aligned}$