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 Taylor series approach suggested by @Matteo and @WillJagy is the way to go.  Here is that approach made a bit more explicit:


*

*Divide the numerator and denominator by x^6.

*Construct the resulting first order Taylor series for both numerator and denominator.

*Observe what the limits must be.


$$\lim_{x\to 0^+} \, \frac{-\tan ^{-1}\left(2 x^3\right)+\frac{(x+2) x^5}{x^2+3}+2 x \sin ^2(x)}{x \exp \left(-\frac{1}{x}\right)-2 x^2+\log \left(\frac{x^2+1}{1-x^2}\right)}$$
$$\lim_{x\to 0^+} \,{ {(\frac{1}{3}-\frac{2 x}{15}+o(x^2) )}\over{(\frac{2}{3}+e^{-1/x}(\frac{1}{x^5}+o(x^2)))}}=\frac{1}{2}$$
Here is a plot of the function:

