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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.

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up vote 3 down vote accepted

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 $\frac 13$.

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ok thanks, looks great. – mathsalomon Aug 25 '11 at 22:08
@anon: you only have to compute the first derivative of the integrand, since we are interested in $f^{(3)}$ (we will know $f''$). – Davide Giraudo Aug 25 '11 at 22:12

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?
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Brillant!! thanks! – mathsalomon Aug 25 '11 at 22:19
This is a simple answer. – leo Aug 26 '11 at 0:47

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

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This is just what I thought when reading the problem. – robjohn Aug 26 '11 at 0:29

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.

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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.

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Using L'Hopital's rule just once, followed by some trivial algebraic simplification, works. Differentiating the numerator gets rid of the "$\int$".

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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 ${1 \over 3}$.

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