TITLE: Evaluating ${\scriptsize\lim \limits_{x \to 0}} \frac{1}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt}$
QUESTION [5 upvotes]: Could someone give a suggestion to calculate this limit please?
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt}$$
Thanks in advance.

REPLY [4 votes]: 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.

REPLY [3 votes]: Here's a heuristic solution: when $t$ is small, $t^2/(1+t^4) \approx t^2$. So if $x$ is small, $$ \int_0^x t^2/(1+t^4) \: dt \approx \int_0^x t^2 \: dt = x^3/3 $$
and this suggests that your limit is $1/3$.  This can be made rigorous by the methods that have already been mentioned.

REPLY [3 votes]: If you change variables $u = t^3$ in the integral, the limit becomes
$$\lim_{x \rightarrow 0} {1 \over x^3}\int_0^{x^3}{1 \over 3}{1 \over 1 + u^{4 \over 3}}\,du$$
You can replace $x^3$ by $x$ here, so that the limit is the same as
$$\lim_{x \rightarrow 0} {1 \over x}\int_0^{x}{1 \over 3}{1 \over 1 + u^{4 \over 3}}\,du$$
By the fundamental theorem of calculus, this limit is just the integrand evaluated at $u = 0$, namely ${1 \over 3}$.