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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:

$lim_{hto 0} f(a+h)-f(a)= lim_{hto 0} {frac {f(a+h)-f(a)}{h}cdot h}$

$=lim_{hto 0} {frac {f(a+h)-f(a)}{h}cdot lim_{hto 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!

The first statement can be proved in a way you already drafted using the total derivate
$$lim_{hto 0} lVert f(a+h)-f(a)rVert \
= lim_{hto 0} {left( frac {lVert f(a+h)-f(a)rVert}{lVert hrVert}cdot lVert hrVertright) } \
=lim_{hto 0} {frac {lVert f(a+h)-f(a)rVert}{lVert hrVert}cdot lim_{hto 0}lVert hrVert}\
=lVert f'(a)rVertcdot0\
=0$$

$a$ 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

$$ lim_{(h,0)to 0} frac{f(h,0)-f(0,0)}{h}= lim_{hto 0}frac{|h|}{h}=begin{cases}-1, h uparrow 0 \ +1, h downarrow 0 end{cases} $$
so this limit depends on how you approach $(0,0)$.

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.