Question: How do we calculate
$$2{\vec d\cdot \vec m\over \vec m\cdot \vec m}\vec m-\vec d$$

given the data ${\vec d = (1, 0)}$ and ${\vec m = (1, 1)}$?

The answer should be $(0, 1)$, but I don't know why.

Answer: 
Observe
\begin{align}
\frac{(\cos x)^2}{\sin (x^2)} = \frac{(1-\frac{1}{2!}x^2+\frac{1}{4!}x^4+\dots)^2}{x^2-\frac{1}{3!}x^6+\frac{1}{5!}x^{10}-\ldots}=\frac{1-x^2+\ldots}{x^2(1-\frac{1}{3!}x^4+\frac{1}{5!}x^8-\ldots)}
\end{align}
which means 
\begin{align}
\lim\_{x\rightarrow 0}\left(\frac{(\cos x)^2}{\sin (x^2)}-\frac{1}{x^2}\right) = -1.
\end{align}

Edit: Rewrite
\begin{align}
\frac{\cos^2x}{\sin(x^2)}-\frac{1}{x^2} = \frac{1-\sin^2 x}{\sin(x^2)}-\frac{1}{x^2}= \left(\frac{1}{\sin(x^2)}-\frac{1}{x^2}\right)-\frac{\sin^2x}{\sin(x^2)}
\end{align}
then observe
\begin{align}
\lim\_{x\rightarrow 0}\frac{\sin^2 x}{\sin(x^2)}=\lim\_{x\rightarrow 0}\frac{\sin x\cos x}{x\cos(x^2)} = \lim\_{x\rightarrow 0}\frac{\sin x}{x}\cdot \frac{\cos x}{\cos(x^2)} = 1. 
\end{align}
I will leave it as an exercise to show that
\begin{align}
\lim\_{x\rightarrow 0}\left( \frac{1}{\sin(x^2)}-\frac{1}{x^2}\right)=0.
\end{align}
You can simplify your life by noting that 

$$\frac{1}{\sin x}-\frac{1}{x}=x\frac{x}{\sin x}\frac{x-\sin x}{x^3}\to 0$$

So $$\frac{(\cos x)^2}{\sin (x^2)}-\frac{(\cos x)^2}{x^2}\to 0$$

and the limit becomes 
$$\frac{(\cos x)^2}{x^2}-\frac{1}{x^2}$$
 which is much easier, even if you use the hospital rule.