My teacher gave us this problem where we need to find:
$$\lim_{(x,y) \to (0,0)} \frac{1-cos(x+y)}{x+y}$$
My first gut instinct would have been to try sandwich theorem (after attempting a few paths and getting all $0$s). However, he gave us a solution using a Taylor approximation:
$$\lim_{(x,y) \to (0,0)} \frac{\frac1 2 (x^2+2xy+y^2)+R_2(x,y)}{x+y} = \lim_{(x,y) \to (0,0)} \frac{\frac1 2 (x+y)^2}{x+y}+ \frac{R2(x,y)}{x+y}=0$$
It seems he used a second degree Taylor polynomial with its remainder term (which tends to $0$ when the function and the Taylor polynomial are close to the same point).
However, three questions arise:
1) Can I always use this Taylor approach to avoid taking a limit by using paths and sandwich theorem? If not, when exactly can or cannot I use Taylor for limits?
2) Why a second degree polynomial? How should I know which grade to use in this technique?
3) I don't get where the $2xy$ came from in the polynomial (inside the parentheses) and why some terms are positive instead of negative. Since $\cos(x+y)$ equals $1$ when evaluated at $(0;0)$, the first derivatives evaluated at $(0;0)$ are equal to $0$ and the second derivatives are all equal to $-1$ then, to me, the polynomial should look like: $$ 1+\frac 1 2 (-x^2-xy-y^2)+R_2(x,y) $$
This limit can be computed as follows $$\lim_{(\vec{r}\cdot\vec{u})\to 0}\frac{1-\cos{(\vec{r}\cdot\vec{u})}}{(\vec{r}\cdot\vec{u})}\tag1$$ With $\vec{u}$ an arbitrary unitary vector and $\vec{r}$ the position vector both of them 2 dimensional. Then you can expand the $\cos{(\vec{r}\cdot\vec{u})}$ in Taylor series $$\cos{(\vec{r}\cdot\vec{u})}\approx 1-\frac{1}{2}(\vec{r}\cdot\vec{u})^2+\mathcal{o}((\vec{r}\cdot\vec{u})^2)$$ Substituting back in $(1)$ $$\lim_{(\vec{r}\cdot\vec{u})\to 0}\frac{1-\cos{(\vec{r}\cdot\vec{u})}}{(\vec{r}\cdot\vec{u})}=\lim_{(\vec{r}\cdot\vec{u})\to 0}\left[(\vec{r}\cdot\vec{u})+\mathcal{o}((\vec{r}\cdot\vec{u}))\right]=\lim_{r\to 0}r\cos{\theta}=0$$ whatever $\theta$ we take