While doing my usual homework-routine I came across a problem which I just can't seem to get my head around, probably missing some fundamental knowledge on the subject, or just not thinking straight.
Anyhow, the exercise is to evaluate the limit as x->0 of the function $\frac{(ln(1+x)-sin(x)-cos(x)+1)}{arctan(x)-x}$ using Maclaurin-expansion for each of the terms. That I have done, but I can't seem to get anywhere after some of the first terms in each series cancel out. I am left with one power series plus another, divided by a third... I know the answer is supposed to be $-3\over 2$.
If anyone would be so kind as to at least point me in the right direction, that would be awesome.
$$\frac{\ln(1+x)-\sin{x}-\cos{x}+1}{\arctan{x}-x}=$$ $$=\tfrac{\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}...\right)-\left(x-\frac{x^3}{6}+\frac{x^5}{120}...\right)-\left(1-\frac{x^2}{2}+\frac{x^4}{24}+...\right)+1}{x-\frac{x^3}{3}+\frac{x^5}{5}-...-x}=$$ $$=\tfrac{\left(x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)\right)-\left(x-\frac{x^3}{6}+o(x^3)\right)-\left(1-\frac{x^2}{2}+o(x^3)\right)+1}{x-\frac{x^3}{3}+o(x^3)-x}=$$ $$=\frac{\frac{x^3}{3}+\frac{x^3}{6}+o(x^3)}{-\frac{x^3}{3}+o(x^3)}=\frac{\frac{1}{2}+\frac{o(x^3)}{x^3}}{-\frac{1}{3}+\frac{o(x^3)}{x^3}}\rightarrow-\frac{3}{2}.$$