Calculus continuity question. show that the function
f(x,y)= |x-1| + |y-1| is continuous at (2,2)
using epsilon delta definition.
The way I have done this is as follows.
|f(x,y)-f(2,2)
= ||x-1|+|y-1|-(1+1)|
= ||x-1|+|y-1|-2|
<= ||x-1|| + ||y-1|| + |2|
<= sqrt( (x-1)^2+(y-1)^2) + sqrt((x-1)^2+(y-1)^2)+2
<= 2sqrt( (x-1)^2+(y-1)^2)+2
 <2*e/2 +2=e+2
 A: You want to prove that when $(x,y) \, \rightarrow \, (2,2)$, then $f(x,y) \, \rightarrow \, f(2,2) = 2$.
For $(x,y) \in \mathbb{R}^{2}$, you have :
$$ 
\begin{eqnarray*}
\vert f(x,y) - f(2,2) \vert & = & \Big\vert \vert x-1 \vert + \vert y-1 \vert - 2 \Big\vert \\
& = & \Big\vert \left( \vert x-1 \vert - 1 \right) + \left( \vert y-1 \vert - 1 \right) \Big\vert \\
& \leq & \Big\vert \vert x-1 \vert - 1 \Big\vert + \Big\vert \vert y-1 \vert - 1 \Big\vert \\
\end{eqnarray*}
$$
By triangular inequality, you have $\Big\vert \vert x-1 \vert - 1 \Big\vert \leq \vert x-2 \vert$ and $\Big\vert \vert y-1 \vert - 1 \Big\vert \leq \vert y-2 \vert$. So :
$$ \vert f(x,y) - f(2,2) \vert \leq \vert x-2 \vert + \vert y-2 \vert $$

Let $\varepsilon > 0$ and let $\delta = \frac{\varepsilon}{2} > 0$. Then, the following is true :
$$ \forall (x,y) \in \mathrm{B}\Big( (2,2), \delta \Big), \; \vert f(x,y) - f(2,2) \vert \leq \varepsilon $$

where $\mathrm{B}\Big( (2,2),\delta \Big) = \left\{ (x,y) \in \mathbb{R}^{2}, \; \Vert (x,y) - (2,2) \Vert \leq \delta \right\}$.
