# Find $\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t$ without using L'hopital's rule

Find $$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t$$ without using L'hopital's rule. Using the rule, I got it as 1/12. What's a way to do it without L'hopital?

A possibility would be to make the change of variable $t=ux$. Then the limit you are looking for is : $$l=\lim_{x \rightarrow 0} \int_0^1 \dfrac{u}{u^4x^4+4} \dfrac{\log(1+ux)}{x} \, du$$

Then, using the dominated convergence theorem, you get the limit inside the integral : $$\begin{array}{rcl} l & = & \displaystyle \int_0^1 \lim_{x \rightarrow 0} \dfrac{u}{u^4x^4+4} \dfrac{\log(1+ux)}{x} \, du\\ & = & \displaystyle \int_0^1 \dfrac{u^2}{4} \, du \\ & = & \dfrac{1}{12} \end{array}$$

Edit : a little bit more of details for the dominated convergence theorem, you can use : $$\dfrac{u}{u^4x^4+4} \dfrac{\log(1+ux)}{x} \leq \frac{u^2}{4}$$ using the classical $\log(1+t) \leq t$.

• I assume you used l'Hospital to compute the limit with the log over $x$, but still this answer is different in flavor than the obvious option :) +1! – Patrick Da Silva Jun 21 '15 at 22:07
• Well I used the fact that $\log(1+t)/t$ tends to $1$ when $t$ goes to 0, and this comes only from the fact that $\log'(1)=1$. – Sylvain L. Jun 21 '15 at 22:13
• That's precisely what I meant, but as I said in the other answer, L'Hospital's rule, Taylor expansion, $o(x^n)$-approximations, they're all one and the same. But the idea at the beginning here is different. – Patrick Da Silva Jun 22 '15 at 0:19

Integration by parts is a reasonable choice. Since: $$\int \frac{t}{t^4+4}\,dt = \frac{1}{4}\arctan\left(\frac{t^2}{2}\right)\tag{1}$$ we have: $$\int_{0}^{x}\frac{t\log(1+t)}{t^4+4}=\frac{\log(1+x)}{4}\arctan\left(\frac{x^2}{2}\right)-\frac{1}{4}\int_{0}^{x}\frac{\arctan\frac{t^2}{2}}{1+t}\,dt \tag{2}$$ and by considering the Taylor series of $\log(1+x)$ and $\arctan\left(\frac{x^2}{2}\right)$ in a neighbourhood of the origin we easily get: $$\lim_{x\to 0}\frac{1}{x^3} \int_{0}^{x}\frac{t\log(1+t)}{t^4+4}=\frac{1}{4}\cdot\frac{1}{2}-\frac{1}{4}\cdot\frac{1}{6}=\color{red}{\frac{1}{12}}\tag{3}$$ as wanted.

As pointed by Patrick Da Silva in the comments, we can also skip the integration-by-parts step, since we are just integrating $\frac{t^2}{4}+o(t^2)$ between $0$ and $x$, hence we trivially have $\frac{x^3}{12}+o(x^3)$.

• I think it's just easier to look at the Taylor expansion of $\log(1+x)$ around the origin without integrating by parts and do the exact same trick. It almost amounts to using l'Hospital's rule, but that's because anything which uses the derivatives amounts to the same trick : Hospital, Taylor expansions, $o(x^n)$-approximations... – Patrick Da Silva Jun 21 '15 at 21:35
• @PatrickDaSilva: yes, you're right, we are just integrating $\frac{t^2}{4}+o(t^2)$ so it is not difficult to compute the limit also without integration by parts. Good point. – Jack D'Aurizio Jun 21 '15 at 21:38
• @PatrickDaSilva: but you are not integrating $t\ln (t+1)$ rather $\frac{t\ln (1+t)}{t^4+4}$ so when you expand the logarithm you still have to integrate by parts several times I think. – Matematleta Jun 21 '15 at 21:58
• @Chilango : You can expand $1/(t^4 + 4)$ using a geometric series argument and expand $\log(1+t)$ using the classical trick, there's no issue there. – Patrick Da Silva Jun 21 '15 at 22:00
• @PatrickDaSilva :Of course. Nice. Thanks! – Matematleta Jun 21 '15 at 22:17

Let's first analyze the limit when $x \to 0^{+}$. If $0 < t < 1$ then we know that $$t - t^{2} = t(1 - t) < \frac{t}{1 + t} < \log(1 + t) < t$$ Multiplying by $t$ we get $$t^{2} - t^{3} < t\log(1 + t) < t^{2}\tag{1}$$ Also we can easily see that $$\frac{1}{4}\left(1 - \frac{t}{4}\right) < \frac{1}{t + 4} < \frac{1}{t^{4} + 4} < \frac{1}{4}\tag{2}$$ If $0 < t < 1$ then all the terms involved in the inequalities $(1)$ and $(2)$ are positive and hence we can multiply them together to get $$\left(t^{2} - t^{3}\right)\left(\frac{1}{4} - \frac{t}{16}\right) < \frac{t\log(1 + t)}{t^{4} + 4} < \frac{t^{2}}{4}$$ or $$\frac{t^{2}}{4} + o(t^{2}) < \frac{t\log(1 + t)}{t^{4} + 4} < \frac{t^{2}}{4}$$ Integrating the above between $0$ and $x$ where $0 < x < 1$ we get $$\frac{x^{3}}{12} + o(x^{3}) < \int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt < \frac{x^{3}}{12}$$ and dividing by $x^{3}$ we get $$\frac{1}{12} + o(1) < \frac{1}{x^{3}}\int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt < \frac{1}{12}$$ and thus by Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{1}{x^{3}}\int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt = \frac{1}{12}$$ A similar argument can be made for the case when $x \to 0^{-}$.

We don't need to use any advanced tools like Taylor series or L'Hospital's Rule and job is done via very simple inequalities and Squeeze theorem.

$$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t^2}{t^2+4}\cdot \underbrace{\frac{\ln (1+t)}{t}}_{\text{approaches 1 when x\to0}}\cdot \underbrace{\frac{t^2+4}{t^4+4}}_{\text{approaches 1 when x\to0}}\mathrm{d}t$$ $$=\lim_{x\to 0} \frac{x-2 \tan ^{-1}\left(x/2\right)}{x^3}=\lim_{x\to0} 1/12 - x^2/80+\mathcal{O}(x^4)=1/12.$$