Basic questions in Real Analysis I started studying Real Analysis 3 and stumbled on the new definition of differentiability, approximation by line/plane. It seems easy to digest but I have some questions about it.
Is $f(x,y)=x$ differentiable at $(0,0)$?
I think it should be differentiable but I keep getting, it's not differentiable:
$f(h_a,h_b)=f(0,0)+\alpha0+\beta0+g(0,0,h_a,h_b)$ gives that
$g(0,0,h_a,h_b)=h_a$ and $g(0,0,h_a,h_b)$ divided by norm of $h$ approaches $\frac{1}{\sqrt{2}}$ if I take $h_a=h_b$.
Here, $g(a,b,h_a,h_b)$ is the error term and $f(a,b)+\alpha h_a+\beta h_b$ represents my plane for approximating the value of $f(a+h_a,b+h_b)$.
Is it not differentiable? --- This has been answered.
Another question, let's say the partial derivative of a function w.r.t. $x$ and $y$ is $0$. Based on my knowledge and physics background, I see that function is not changing along the $x$ and $y$-axis. Doesn't it imply our particle at origin is not moving at all? If it changes in any direction then it must have components along the $x$ and $y$-axis but there are no changes in those directions. Doesn't it imply that the function must be differential at the origin if partial derivatives are $0$?
I know, there exist some continuous functions with partial derivatives $0$ at origin and are non-differentiable.
 A: For $f(x, y)=x$, we have
$$f(h_a, h_b) = f(0, 0) + 1 \cdot h_a + 0 \cdot h_b + \underbrace{g(0, 0, h_a, h_b)}_{=0}.$$
As noted in the comments, the original function $f$ is linear, so the linear approximation has zero error.
If you choose the "wrong" plane for the linear approximation, you will of course get an error term that doesn't vanish at the desired rate.
A: A function $f:\Omega(\subseteq\Bbb R^n)\to\Bbb R$ is (everywhere) differentiable if the limit
$$\lim_{\boldsymbol \epsilon\to 0}\frac{f(\boldsymbol x+\boldsymbol \epsilon)-f(\boldsymbol x)}{|\boldsymbol \epsilon|}$$
Exists $\forall \boldsymbol x\in \Omega$.  In the case of $f\big((x_1,x_2)\big)=x_1$, you can write
$$f(\boldsymbol x)=\boldsymbol x\cdot(1,0)$$
Hence
$$\lim_{\boldsymbol \epsilon\to 0}\frac{f(\boldsymbol x+\boldsymbol \epsilon)-f(\boldsymbol x)}{|\boldsymbol \epsilon|}=\lim_{\boldsymbol \epsilon\to 0}\frac{(\boldsymbol x+\boldsymbol \epsilon)\cdot(1,0)-\boldsymbol x\cdot (1,0)}{|\boldsymbol \epsilon|}$$
Using the linearity of the dot product,
$$=\lim_{\boldsymbol \epsilon\to 0}\frac{\boldsymbol x\cdot (1,0)-\boldsymbol x\cdot(1,0)+\boldsymbol \epsilon\cdot (1,0)}{|\boldsymbol \epsilon|}=\lim_{\boldsymbol \epsilon\to 0}\frac{\boldsymbol \epsilon\cdot (1,0)}{|\boldsymbol \epsilon|}$$
And since $\boldsymbol \epsilon\cdot (1,0)\leq |\boldsymbol \epsilon|$ this goes to zero.
