I have to solve the following exercise:
Give $a$, $b$, $c \in \mathbb R$ such that $$\frac{1}{1-\cos x} = \frac{a}{x^2} + b +cx^2 + o(x^2)$$ for $x\to 0$.
Here's my attempt:
I know that $$\cos x = \frac{1}{2}\left( e^{ix} + e^{-ix}\right) = \frac{1}{2} \sum_{n=0}^\infty \frac{(ix)^{2n}}{(2n)!} = \frac{1}{2}\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!} = \frac{1}{2}\left(1-\frac{x^2}{2}+o(x^2)\right)$$
for $x\to 0$.
The question now is how to resolve $\frac{1}{1-\cos x}$. This much looks like an application of the geometric series here. I assume I can use it since $\cos x < 1$ for all sufficiently small $x \neq 0$. I wouldn't know how to solve this exercise otherwise. Proceeding, it follows that
$$\frac{1}{1-\cos x} = \frac{1}{1-\left(\frac{1}{2} - \frac{x^2}{4} + o(x^2)\right)} = \sum_{n=0}^\infty\left(\frac{1}{2} - \frac{1}{4}x^2+o(x^2)\right)^n = 1+ \frac{1}{2}-\frac{1}{4}x^2+o(x^2).$$
Choosing $a=0$, $b = 1,5$ and $c=-\frac{1}{4}$ establishes the case.
Your expansion of $\cos(x)$ at $0$ is obviously wrong, since it implies $\lim_{x\to 0}\cos(x)=\frac 12$.
You needn't use the power series for $e^{ix}$. I'd suggest that you rather use Taylor expansion at order $6$ for $\cos$, since the derivatives at $0$ are very simple. This method yields $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6)$$
Hence $$\begin{align} \frac{1}{1-\cos x} &=\frac{1}{1-\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6) \right)}\\ &=\frac{2}{x^2}\frac{1}{1-\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)}\\ &=\frac{2}{x^2} \left( 1+\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)+ \left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)^2+ o\left(\left(\frac{x^2}{12}-\frac{x^4}{360}+o(x^4)\right)^2\right)\right)\\ &=\frac{2}{x^2}\left( 1+\frac{x^2}{12}-\frac{x^4}{360}+\frac{x^4}{144}+o\left(x^4\right)\right)\\ &=\frac{2}{x^2} + \frac 16 + \frac{x^2}{120}+o(x^2) \end{align}$$