Evaluating limit of multi variables The question:
$$\lim_{(x,y)\rightarrow(0,0)}\frac{\sin(x+y)}{x+y}.$$
I tried this using $y=x$ path and $y=0$ path and both approach to the same value, $1$. My problem is to show that this value really approaches to one on any path 
using the $\varepsilon$-$\delta$ definition. 
 A: Consider the function :
$$
f:
  \begin{array}{rcl}
    \mathbb{R}^{2} & \longrightarrow &\mathbb{R} \\
    (x,y) & \longmapsto & \sin(x+y) \\
  \end{array}
$$
For $(x,y)$ in a neighborhood of $(0,0)$ in $\mathbb{R}^{2}$, a Taylor expansion of $f$ gives :
$$ f\big( (x,y) \big) = f\big( (0,0) \big) + \left\langle \nabla f \big( (0,0) \big) , \begin{pmatrix} x \\ y \end{pmatrix} \right\rangle_{\mathbb{R}^{2}} + o\big( \Vert (x,y) \Vert \big). $$
(where $\left\langle \cdot,\cdot \right\rangle_{\mathbb{R}^{2}}$ is the usual inner product in $\mathbb{R}^{2}$, namely : $\left\langle u,v \right\rangle_{\mathbb{R}^{2}} = u^{\top}v$). 
Here, 
$$ \nabla f\big( (0,0) \big) = \begin{pmatrix} \cos(0) \\ \cos(0) \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}. $$
Therefore, the Taylor expansion around $(0,0)$ writes :
$$ \sin(x+y) = f\big( (x,y) \big) = (x+y) + o\big( \Vert (x,y) \Vert \big). $$
It follows that :

$$ \frac{\sin(x+y)}{x+y} \, \mathop{\longrightarrow} \limits_{\Vert (x,y) \Vert \to 0} \; 1. $$

A: Just set $z=x+y$ and use the single valued case to show that $\frac{\sin z}{z}\rightarrow 1$ as $z\rightarrow 0$.
