$\displaystyle \lim_{x\to0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1} $
Really drawing a blank on this one.
The function $\displaystyle \frac{t^2}{t^4+1}$ is defined and continuous everywhere so that integral exists and it's a differentiable function of x. Let's call F the primitive function of the integrand, i.e. $\displaystyle F'(x)=\frac{x^2}{x^4+1}$, and then according to the Fundamental Theorem of Integral Calculus and applying L'Hospital's rule (why can we?):
$\displaystyle \lim_{x\to0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1} =\lim_{x\to 0}\frac{F(x)-F(0)}{x^3}=\lim_{x\to0}\frac{F'(x)}{3x^2}=\lim_{x\ to0}\frac{1}{3(x^4+1)}=\frac{1}{3}$
And let us thank the gods, godesses and nymphs of the rivers that we can use the above trick, because if you wanted to evaluate the integral you'd get (sit well!):
$\displaystyle \int\frac{t^2}{t^4+1}dt=-\frac{1}{4\sqrt{2}}\,\ln\!\!\left(\frac{t^2+\sqrt{ 2}t+1}{t^2-\sqrt{2}t+1}\right)+\frac{1}{2\sqrt{2}}\,\arctan\! \!\left(\frac{\sqrt{2}t}{1-t^2}\right)+C(onstant)$
Of course, it's done by partial fractions: $\displaystyle t^4+1=(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)$
Tonio