If $f$ is differentiable, then $f$ is continuous Let  $f: \mathbb{R}^2 \to \mathbb{R} $ be a function. Prove or disprove:
1) If $f$ is differentiable at $(a,b)$, then $f$ is continuous at $(a,b)$ 
2) If $f$ is continuous at $(a,b)$, then $f$ is differentiable at $(a,b)$ 
What I already have:
If I want to show that $f$ is differentiable at $a$ (and with that also continuous  at $a$), I do it like this:
$\lim_{h\to 0}  f(a+h)-f(a)= \lim_{h\to 0} {\frac {f(a+h)-f(a)}{h}\cdot h}$
$=\lim_{h\to 0} {\frac {f(a+h)-f(a)}{h}\cdot \lim_{h\to 0}h}=f'(a)\cdot0=0 $
However here I need to show it for a point $(a,b)$. My idea was to show it like this:
If $f$  is differentiable at $(a,b) \to \forall h \in \mathbb{R}^2$
$f((a, b) + h) - f(a, b) = \nabla f(a,b)*h + r(h)$ and $\frac{r(h)}{|h|  }  \to 0 $  for     $h \to 0$ 
2) I know that this is not true. So would a counterexample be $f(x,y)= |x| + y| ? $
Thanks in advance! 
 A: The first statement can be proved in a way you already drafted using the total derivate
$$\lim_{h\to 0}  \lVert f(a+h)-f(a)\rVert \\
= \lim_{h\to 0} {\left( \frac {\lVert f(a+h)-f(a)\rVert}{\lVert h\rVert}\cdot \lVert h\rVert\right) } \\ 
=\lim_{h\to 0} {\frac {\lVert f(a+h)-f(a)\rVert}{\lVert h\rVert}\cdot \lim_{h\to 0}\lVert h\rVert}\\
=\lVert f'(a)\rVert\cdot0\\
=0$$
$a$ is a vector from $\mathbb{R^2}$ and $f'(a)$ is a linear mapping from $\mathbb{R^2}\to \mathbb{R}$.
The second statement can be disproven by your counterexample.
The function
$$f(x,y)= |x| + |y| $$
is continous, but at $(0,0)$ we have
$$ \lim_{(h,0)\to 0}  \frac{f(h,0)-f(0,0)}{h}= \lim_{h\to 0}\frac{|h|}{h}=\begin{cases}-1, h \uparrow 0 \\ +1, h \downarrow 0 \end{cases} $$
so this limit depends on how you approach $(0,0)$.
A: For a function $f : \mathbb{R}^2 \to \mathbb{R}$ (or more generally, from $U \to \mathbb{R}^m$, where $U \subseteq \mathbb{R}^n$ is open), we say that $f$ is differentiable at $\mathbf{x} \in \mathbb{R}^2$ if there exists some linear mapping $A : \mathbb{R}^2 \to \mathbb{R}$ such that
$$\lim_{\mathbf{h} \to \mathbf{0}} \frac{f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x}) - A(\mathbf{h})}{||\mathbf{h}||} = 0,$$ or equivalently
$$f(\mathbf{x} + \mathbf{h}) = f(\mathbf{x}) + A(\mathbf{h}) + o(\mathbf{h}),$$
where $o(\mathbf{h})$ is a function that satisfies $\lim_{\mathbf{h} \to \mathbf{0}} o(\mathbf{h})/||\mathbf{h}|| = 0$. When $A$ exists, we denote it $f'(\mathbf{x})$ or $D_f(\mathbf{x})$. [See https://en.wikipedia.org/wiki/Derivative#Total_derivative.2C_total_differential_and_Jacobian_matrix for further details.]
(To illustrate why we cannot just use 
$$\text{The derivative of $f$ at $\mathbf{x}$ is $f'(\mathbf{x}) = \lim_{\mathbf{h} \to \mathbf{0}} \frac{f(\mathbf{x} + \mathbf{h}) - f(\mathbf{x})}{||\mathbf{h}||}$ when the limit exists,}$$ consider the the function $f((x,y)) = x$. Then $f((x,y) + (u, v)) =  x + u = f((x, y)) + (1,0)\cdot(u,v)$, so is (as expected) differentiable. However, $\lim_{u \to 0} (f((x,y)+(u,0)) - f((x,y)))/||(u,0)|| = \lim_{u \to 0} u/|u|$, which does not exists.)
We can however use the second definition to prove that a function differentiable at a point is also continuous at that point.
1) We have
$$0 \leq \lim_{\mathbf{h} \to \mathbf{0}} |f((a,b) + \mathbf{h}) - f((a, b))| = \lim_{\mathbf{h} \to \mathbf{0}} |A(\mathbf{h}) + o(\mathbf{h})| \leq \lim_{\mathbf{h} \to \mathbf{0}} (||A|| + 1)||\mathbf{h}|| = 0,$$
where the inequality is deduced from $|A(\mathbf{h})| \leq ||A||||\mathbf{h}||$ and from $o(\mathbf{h})/||\mathbf{h}|| \to 0$, which implies eventually $|o(\mathbf{h})/||\mathbf{h}||| \leq 1 \implies |o(\mathbf{h})| \leq ||\mathbf{h}||$. Hence, we must have equality, and therefore $f$ is continuous at $(a,b)$.
2) The proof given by @miracle173 is correct, but instead because differentiability in $\mathbb{R}^2$ implies the existence of directional/partial derivatives. Note also that $f((x,y))=|x|$ would have sufficed.
