How find this limit $\lim_{x\to 0}\frac{\int_{0}^{x}\sin{t}\ln{(1+t)}dt-\frac{x^3}{3}+\frac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}$ 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?
 A: 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.
A: 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.
$\begin{align}
I& =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\sin{t}\ln{(1+t)}\,dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{(x-\sin{x})(e^{x^2}-1)}\\
\\
& =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\left(t-\dfrac{t^3}{6}+\dfrac{t^5}{120}+O(t^7)\right)\cdot\left(t-\dfrac{t^2}{2}+\dfrac{t^3}{3}-\dfrac{t^4}{4}+O(t^5)\right)dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\left(\dfrac{x^3}{6}-\dfrac{x^5}{120}+O(x^7)\right)\left(x^2+\dfrac{x^4}{2}+O(x^6)\right)}\\
\\
& =\lim_{x\to 0}\dfrac{\displaystyle\int_{0}^{x}\left(t^2-\dfrac{t^3}{2}+\dfrac{t^4}{6}+O(t^5)\right)dt-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\left(\dfrac{x^3}{6}-\dfrac{x^5}{120}+O(x^7)\right)\left(x^2+\dfrac{x^4}{2}+O(x^6)\right)}\\
\\
& =\lim_{x\to 0}\dfrac{\dfrac{x^3}{3}-\dfrac{x^4}{8}+\dfrac{x^5}{30}+O(x^6)-\dfrac{x^3}{3}+\dfrac{x^4}{8}}{\dfrac{x^5}{6}+O(x^7)}\\
\\
&= \lim_{x\to 0}\dfrac{\dfrac{x^5}{30}+O(x^6)}{\dfrac{x^5}{6}+O(x^7)}\\
&= \dfrac{1}{5}
\end{align}$
A: 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. 
