Evaluating ${\scriptsize\lim \limits_{x \to 0}} \frac{1}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt}$

Answer the following:

• If $\displaystyle F(x) = \int_{0}^{x} f(t) \text{ d}t$, and $\displaystyle f$ is continuous, then what it $\displaystyle F'(x)$ (the derivative of $\displaystyle F$)?
• What is $\displaystyle \lim_{x \to 0} \int_{0}^{x} f(t) \text{ d}t$ for continuous $\displaystyle f$?
• Do you know L'Hopital's rule?

Suggestion: use the substitution $u=t/x$, then just cancel and plug in $x=0$.

One can do this in an elementary and explicit way: write $\dfrac{t^2}{1+t^4}=t^2-A(t)$ with $A(t)=\dfrac{t^6}{1+t^4}$ hence $0\le A(t)\le t^6$. Integrating this from $t=0$ to $t=x$ and dividing by $x^3$ yields that the ratio $R(x)$ one is interested in satisfies $$ \frac13-\frac17x^4\le R(x)\le\frac13. $$ From this double inequality, the limit of $R(x)$ when $x\to0$ should be clear.