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? 
 A: 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 $1/2$
A: You made a mistake by considering only the first asymptotic term of the algebraic fraction. Be careful:
$$\frac{2x^5+x^6}{3+x^2}=\frac{2}{3}x^5 + \frac{1}{3}x^6 + o(x^6).$$
Then your limit gives $\frac{1}{2}$, as results from the comment by JimB. 
A: 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:

