# Evaluate $\lim\limits_{x\to 0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}dt$ . [closed]

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

To determine the above value, do I need to integrate first? Or, is there some other way to do this?

## closed as off-topic by Xander Henderson, mrtaurho, Cesareo, Gibbs, José Carlos SantosFeb 18 at 10:22

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• Try L'Hopital's rule. – Antonio Vargas Jun 29 '17 at 13:00
• See my edit for the answer to your third question (if we count "or is there any other method to do?" as the second). For your first question, I think the idea is to use l'Hopital's rule, which means you'll need to use the FTC to differentiate $\displaystyle\int_0^x \frac{t^2}{t^4+1} \, dt$ with respect to $x$. So, no, you don't actually integrate it. – tilper Jun 29 '17 at 13:02
• Yes${}{}{}{}{}$ – tilper Jun 29 '17 at 13:05
• The intended method here is surely l'Hopital's rule, but one can also expand the integrand in a Taylor series: $t^2 (1 - t^4 + O(t^5)) = t^2 + O(t^3)$. Then, integrating gives that the integral is $\frac{1}{3} x^3 + O(x^4)$, so that the argument of the limit is $\frac{1}{3} + O(x)$. – Travis Jun 29 '17 at 13:20
• @Beverlie You're welcome. Using power series is very often an efficient method for computing certain classes of limits, and the method sometimes shines in cases in which l'Hopital is impractical, e.g., limits $\lim_{x \to 0} \frac{f(x)}{g(x)}$ where $f$ and $g$ are at least modestly complicated functions with zeros of some high order at $x = 0$. – Travis Jun 29 '17 at 13:27

This is an application of L'Hôpital's Rule because $$\lim_{x\to 0}\int_0^x\frac{t^2}{t^4+1}dt=0$$ and $$\lim_{x\to 0}x^3=0.$$ So, we have: $$\lim_{x\to 0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}dt=\lim_{x\to 0}\frac{\int_0^x \frac{t^2}{t^4+1}dt}{x^3}\stackrel{\text{LH}}{=}\lim_{x\to 0}\frac{x^2}{3x^2(x^4+1)}=\lim_{x\to 0}\frac{1}{3(x^4+1)}=\frac{1}{3}.$$

Let $f(x)=\int_0^x\dfrac{t^2}{t^4+1}dt$

Now you have to compute $\lim\limits_{x\to 0}\dfrac{f(x)}{x^3}$.

This can be done in a hospital !

• Check your spelling: "This can be done in a hospital" – Toby Mak Jun 29 '17 at 13:04
• sense of humor.. thx – Beverlie Jun 29 '17 at 13:04
• @Toby: I checked my spelling, but you should check your humor ! – Fred Jun 29 '17 at 13:06
• Anyone who has taken intermediate-level calculus can still think of the answer in a hospital - the problem is writing it down when they're injured – Toby Mak Jun 29 '17 at 13:06
• @tobymark Sharp critic – Beverlie Jun 29 '17 at 13:09

Alternatively, expand the integrand to Taylor series:

$$\lim\limits_{x\to 0}\frac{1}{x^3}\int_0^x\frac{t^2}{t^4+1}dt=$$ $$\lim\limits_{x\to 0}\frac{1}{x^3}\int_0^x t^2(1-t^4+t^8-...)dt=$$ $$\lim\limits_{x\to 0}\frac{\frac{x^3}{3}-\frac{x^7}{7}+O(x^{11})}{x^3}=\frac13.$$

The integrand $f(t)=\frac{t^2}{1+t^4}$ satisfies the inequality $(1-\epsilon)(t^2)\le f(t)\le t^2$ for $0\le t \le x$ when $x$ is sufficiently small and positive, with $\epsilon$ any small positive number. Therefore by comparison we find that the full expression in the problem lies between $\frac{1-\epsilon}{3}$ and $1/3$ forcing the limit to be $1/3$ as $x$ approaches zero from above. A similar comparison also gives the limit $1/3$ as $x$ approaches zero from below making the limit well defined and equal to $1/3$.

Note that the function under limit is an even function and hence it is sufficient to consider the limit as $x\to 0^{+}$. So in what follows we assume that $x$ is positive.

Note that if $0\leq t\leq x$ then we have $$\frac{t^{2}}{x^{4}+1}\leq\frac{t^{2}}{t^{4}+1}\leq t^{2}$$ and integrating the above with respect to $t$ in interval $[0,x]$ and dividing the resulting inequality by $x^{3}$ we get $$\frac{1}{3(x^{4}+1)}\leq\frac{1}{x^{3}}\int_{0}^{x}\frac{t^{2}}{t^{4}+1}\,dt\leq\frac{1}{3}$$ and now applying Squeeze theorem we get the desired limit as $1/3$.