Find the value of: $$\lim_{x \to 0}\frac{1}{x^3}\int^x_0\frac{t\ln(1+t)}{t^4 + 4}$$

How should I approach this question? I'm not sure that solving the integration first, and then proceeding to deal with the limit would be the best approach. Is there some way to solve it directly?


What you need to care about is the behaviour of the integrand for $t$ close to zero. Therefore, find the first nonzero term in the Taylor expansion: $$ \frac{t \log{(1+t)}}{t^4+4} = t(t+o(t))\left(\frac{1}{4}+o(t^2)\right) = \frac{t^2}{4}+o(t^2) \text{ as } t \to 0. $$ Integrating this preserves the order term: $$ \int_0^x \frac{t \log{(1+t)}}{t^4+4} \, dt = \frac{1}{12}x^3+o(x^3) \text{ as } x \to 0, $$ and then dividing by $x^3$ implies $$ \frac{1}{x^3}\int_0^x \frac{t \log{(1+t)}}{t^4+4} \, dt = \frac{1}{12} + o(1) \text{ as } x \to 0, $$ so the limit has value $1/12$.


Using L'Hospital, given limit is $$\lim_{x\to 0}\frac{x\ln(1+x)}{3x^2(4+x^4)}=\frac{1}{12}$$

  • $\begingroup$ Ohhh....... And why don't we apply product rule in this case? $\endgroup$ – Gummy bears Oct 8 '15 at 13:35
  • $\begingroup$ @Gummybears: What has the product rule got to do with this? $\endgroup$ – user21820 Oct 8 '15 at 13:36
  • $\begingroup$ I basically used L'Hospital in the following form, $$\lim_{x\to 0}\frac{f(x)}{g(x)}=\lim_{x\to 0}\frac{f'(x)}{g'(x)}$$ $\endgroup$ – Samrat Mukhopadhyay Oct 8 '15 at 13:36
  • $\begingroup$ Oh... Yes, I get it. My fault. $\endgroup$ – Gummy bears Oct 8 '15 at 13:39

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