calculus - How to evaluate the $limlimits_{xto 0}frac {2sin(x)-arctan(x)-xcos(x^2)}{x^5}$, using power series?

How to evaluate the $\displaystyle\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, using power series?

It made sense to first try and build the numerator using power series that are commonly used:

$\displaystyle2\sin(x)=\sum_{k=0}^\infty \dfrac{2(-1)^kx^{2k+1}}{2k+1!} = 2x -\frac{x^3}{3}+\frac{x^6}{60} + \dotsb$

$\displaystyle-\arctan(x)=\sum_{k=0}^\infty \dfrac{(-1)^{k+1}x^{2k+1}}{2k+1} = -x +\frac{x^3}{3}-\frac{x^6}{6} + \dotsb$

$\displaystyle-x\cos(x^2)=\sum_{k=0}^\infty \dfrac{(-1)^{k+1}x^{4k+1}}{2k!} = -x +\frac{x^5}{2}+ \dotsb$

Hence,

$\displaystyle\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5} = \lim\limits_{x\to 0} \dfrac{[2x -\frac{x^3}{3}+\frac{x^6}{60} + \dotsb] + [-x +\frac{x^3}{3}-\frac{x^6}{6} + \dotsb] + [x +\frac{x^5}{2}+ \dotsb]} {x^5}$

In similar problems, the there is an easy way to take out a common factor that would cancel out with the denominator, resulting in an easy-to-calculate limit. Here, however, if we were to take a common factor from the numerator, say, $x^6$, then we would end up with an extra $x$

What possible strategies are there to solve this question?