Evaluating ${\scriptsize\lim \limits_{x \to 0}} \frac{1}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt}$ 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.
 A: Suggestion: use the substitution $u=t/x$, then just cancel and plug in $x=0$.
A: 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.
A: Put $\displaystyle f(x):=\int_0^x\frac{t^2}{1+t^4}dt$. It's a smooth function. 
We have $f(0)=f'(0)=f''(0)$, and in order to apply Taylor's theorem compute $f^{(3)}(0)$. 
We have $f'(x) = \dfrac{x^2}{1+x^4}$ hence $$f''(0)=\lim_{x\to 0}\frac{f'(x)-f'(0)}x=\lim_{x\to 0}\frac x{1+x^4}=0.$$
Since $f''(x) = \dfrac{2x}{1+x^4}-\dfrac{4x^5}{(1+x^4)^2}$, we get 
$$f^{(3)}(0)=\lim_{x\to 0}\dfrac{f''(x)-f''(0)}x= 2.$$
Taylor's theorem gives that $f(x)=f(0)+xf'(0)+\frac{x^2}2f''(0)+\frac{x^3}{6}f^{(3)}(0)+x^3\varepsilon(x) =\frac{x^3}{6}f^{(3)}(0)+x^3\varepsilon(x)$, where $\displaystyle \lim_{x\to 0}\varepsilon(x)=0$. We conclude that the limit we where looking for is $\frac 13$.
A: 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.
A: 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 ${1 \over 3}$.
A: Using L'Hopital's rule just once, followed by some trivial algebraic simplification, works.  Differentiating the numerator gets rid of the "$\int$".
A: 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?

