Find this limit $$I=\lim_{x\to 0}\dfrac{\int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}$$
I think $$I=6\lim_{x\to 0}\dfrac{\int_{0}^{x}(\sin{t}\ln{(1+t)}-t^2+\dfrac{t^3}{2})dt}{x^5}$$
Iuse $\sin{x}=x-\dfrac{x^3}{6}+o(x^3)$ and $$\ln{(1+x)}=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+o(x^3)$$ so $$\sin{x}\ln{(1+x)}=x^2-\dfrac{x^3}{2}+\dfrac{1}{6}x^4+o(x^4)$$ so $$I=\dfrac{1}{5}$$
Have other methods?
Using l'Hôpital's rule doesn't look like an ideal approach. It has to be used five times before it yields a result of $\frac{1}{5}$. (Sending Mathematica after the limit yields the same result).
Those two polynomial terms in the numerator look suspiciously like members of some Taylor expansion, though.
The factors of the integrand and the denominator have power series expansions that converge near $0$. If you work out the first few terms, you get the following.
$\begin{align} I& =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\sin{t}\ln{(1+t)}\,dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}\\ \\ & =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\left(t-\dfrac{t^3}{6}+\dfrac{t^5}{120}+O(t^7)\right)\cdot\left(t-\dfrac{t^2}{2}+\dfrac{t^3}{3}-\dfrac{t^4}{4}+O(t^5)\right)dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\left(\dfrac{x^3}{6}-\dfrac{x^5}{120}+O(x^7)\right)\left(x^2+\dfrac{x^4}{2}+O(x^6)\right)}\\ \\ & =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\left(t^2-\dfrac{t^3}{2}+\dfrac{t^4}{6}+O(t^5)\right)dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\left(\dfrac{x^3}{6}-\dfrac{x^5}{120}+O(x^7)\right)\left(x^2+\dfrac{x^4}{2}+O(x^6)\right)}\\ \\ & =\lim_{x\to 0}\dfrac{\dfrac{x^3}{3}-\dfrac{x^4}{8}+\dfrac{x^5}{30}+O(x^6)-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\dfrac{x^5}{6}+O(x^7)}\\ \\ &= \lim_{x\to 0}\dfrac{\dfrac{x^5}{30}+O(x^6)}{\dfrac{x^5}{6}+O(x^7)}\\ &= \dfrac{1}{5} \end{align}$
We can apply LHR as follows. First we note that $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\text{ (via LHR)},\text{ and }\lim_{x \to 0}\frac{e^{x^{2}} - 1}{x^{2}} = 1$$ and then we can write $$\begin{aligned}L &=\lim_{x\to 0}\dfrac{{\displaystyle \int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}}{(x-\sin{x})(e^{x^2}-1)}\\ &= \lim_{x \to 0}\dfrac{{\displaystyle \int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}}{\dfrac{x - \sin x}{x^{3}}\cdot x^{3}\cdot\dfrac{e^{x^{2}} - 1}{x^{2}}\cdot x^{2}}\\ &= 6\lim_{x \to 0}\dfrac{{\displaystyle \int_{0}^{x}\sin{t}\ln{(1+t)}dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}}{x^{5}}\\ &= 6\lim_{x \to 0}\dfrac{\sin x\log(1 + x) - x^{2} + \dfrac{x^{3}}{2}}{5x^{4}}\text{ (Using L'Hospital's Rule)}\\ &= \frac{3}{5}\cdot\lim_{x \to 0}\dfrac{2\sin x\log(1 + x) - 2x^{2} + x^{3}}{x^{4}}\\ &= \frac{3}{5}\cdot\lim_{x \to 0}\dfrac{2(\sin x - x)\log(1 + x) + 2x\log(1 + x) - 2x^{2} + x^{3}}{x^{4}}\\ &= \frac{3}{5}\cdot\left\{\lim_{x \to 0}\dfrac{2(\sin x - x)}{x^{3}}\cdot\dfrac{\log(1 + x)}{x} + \lim_{x \to 0}\dfrac{2\log(1 + x) - 2x + x^{2}}{x^{3}}\right\}\\ &= \frac{3}{5}\cdot\left(-\frac{1}{3} + \lim_{x \to 0}\dfrac{2\log(1 + x) - 2x + x^{2}}{x^{3}}\right)\\ &= \frac{3}{5}\cdot\left(-\frac{1}{3} + \lim_{x \to 0}\dfrac{\dfrac{2}{1 + x} - 2 + 2x}{3x^{2}}\right)\text{ (Using L'Hospital's Rule)}\\ &= \frac{3}{5}\cdot\left(-\frac{1}{3} + \frac{2}{3}\lim_{x \to 0}\dfrac{1 - (1 - x)(1 + x)}{x^{2}(1 + x)}\right)\\ &= \frac{3}{5}\cdot\left(-\frac{1}{3} + \frac{2}{3}\lim_{x \to 0}\dfrac{1}{1 + x}\right)\\ &= \frac{3}{5}\cdot\frac{1}{3} = \frac{1}{5}\end{aligned}$$ Note that L'Hospital's rule should be used together with various manipulations to simplify limit expressions. Just applying L'Hospital repeatedly without any simplification does not help in most cases.
