How to evaluate the $\lim\limits_{x\to 0}\frac {2\sin(x)-\arctan(x)-x\cos(x^2)}{x^5}$, 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? 
 A: Since the denominator has degree $5$, you just need to stop at $x^5$ in the power series expansions. So
\begin{align}
2\sin x&=2x-\frac{x^3}{3}+\frac{x^5}{60}+o(x^5)\\
-\arctan x&=-x+\frac{x^3}{3}-\frac{x^5}{5}+o(x^5)\\
-x\cos(x^2)&=-x+\frac{x^5}{2}+o(x^5)
\end{align}
Summing up we get
$$
\left(\frac{1}{60}-\frac{1}{5}+\frac{1}{2}\right)x^5+o(x^5)=
\frac{19}{60}x^5+o(x^5)
$$
Note that $o(x^5)$ means some unspecified function (but computable doing the math), with the property that $\lim\limits_{x\to0}\dfrac{o(x^5)}{x^5}=0$, so the sum of two of them is still $o(x^5)$.
Thus the limit is $19/60$.
You can be sure that going beyond $x^5$ would not really matter, because when dividing by $x^5$ the numerator would have summands with a factor $x$ that goes to $0$. If all terms of degree at most $5$ would cancel out, the limit would be $0$.
Note also the small error you have in the expansion of the arctangent.
A: if you add up the expansions you have given us, you get 1/2 +0. You take a common factor of $x^5$ from the numerator and divide through you get 1/2+ x(...) --> 1/2
