Evaluating $\lim_{x\to0^+}\frac{2x(\sin x)^2+\frac{2x^7+x^8}{3x^2+x^4}-\arctan(2x^3)}{\ln(\frac{1+x^2}{1-x^2})-2x^2+xe^{-{1\over x}}}$ To start with, the term $xe^{-{1\over x}}$ can be ignored.Then splitting the term $\ln(\frac{1+x^2}{1-x^2})=\ln(1+x^2)-\ln(1-x^2)$ and expanding both of them up to order 3 we have in the denominator $2x^2+{2x^6\over 3}-2x^2={2x^6\over3}$
As for the numerator $\arctan(2x^3)\sim2x^3$ and ${2x^7+x^8\over 3x^2+x^4}\sim{2\over3}x^5$.So if we expand $\sin x$ up to order 2,then take the square (ignoring higher order terms) and multiply by $2x$ we have $2x(\sin x)^2\sim2x^3-{2\over3}x^5$ which cancels out with the other terms,giving us an indeterminate form.So I think we should have $x^6$ term in the numerator.
But if I expand $\sin x$ up to order 3 and take square, I fail to obtain $x^5$ term(so that when multiplied by $2x$ gives me $x^6$).How can this be done? 

The items below are called Taylor Series. This is a good way to reduce errors, in cases where L'Hospital's method leads to messy derivatives. One may take time and get each series correct. In this case, all the series derive from that for $$ \frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 \cdots, $$ which leads fairly quickly to series for $1/(3+t),$ for $\log(1+t),$ and for $\arctan t$
$$ $$
To confirm the first step below, multiply the right hand side by $x^2 + 3$ and notice the cancellations:
$$  \frac{1}{x^2 + 3} = \frac{1}{3} - \frac{x^2}{9} + \frac{x^4}{27} - \frac{x^6}{81}  + \cdots $$
$$  \frac{1}{x^4 + 3x^2} = \frac{1}{3x^2} - \frac{1}{9} + \frac{x^2}{27} - \frac{x^4}{81}  + \cdots $$
$$  \frac{x^8 + 2x^7}{x^4 + 3x^2} = \frac{2x^5}{3} + \frac{x^6}{3} - \frac{2x^7}{9} - \frac{x^8}{9}  + \cdots $$
$$  2x \sin^2 x = 2 x^3 - \frac{2x^5}{3} + \frac{4x^7}{45} -  \cdots $$
$$  \arctan (2x^3)  = 2 x^3 - \frac{8x^9}{3}  -  \cdots $$
The numerator is $$  \frac{x^6}{3} - \frac{2x^7}{15}   + \cdots  $$
The denominator is $$  \frac{2x^6}{3} + \frac{2x^{10}}{5}   + \cdots  $$
giving limit $1/2$
