<s>[INST] 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?
 [/INST] 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.
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