How did they study the differentiability at (0,0) of this particular function? I understand that they are seeing what happens as (h,k) goes to the origin but I do not get the 3 things that I have marked from one to three in the image below. Namely, where they got the '0' from, where the '3/2' came and how the '4' came to be?
I tried plugging in the f(x,y) into the limit function and trying to simplify but I got to the power of'3/4' instead of '3/2'. Could someone please explain how they got these three numbers? Thank you!


 A: The limit that they are studying is given by
$$\lim_{(h,k)\to (0,0)}\frac{f(h,k)-f(0,0)-f_{x}(0,0)(h-0)-f_{y}(0,0)(k-0)}{||((h-0),(k-0))||},\quad (\star)$$
One of the conditions to guarantee differentiability is that the limit $(\star)$ be $0$.

*

*Since for $h\not=0$, we have $$\frac{f(h,0)-f(0,0)}{h}=\frac{\sin^{2}(h)/h^{1/2}}{h}=\frac{\sin^{2}(h)}{h^{3/2}}\rightarrow0,\quad h\to 0$$
and $$\frac{f(0,h)-f(0,0)}{h}=\frac{\sin^{2}(\sin(h))/h^{1/2}}{h}=\frac{\sin^{2}(\sin(h))}{h^{3/2}}\rightarrow 0,\quad h\to 0$$
So, $f_{x}(0,0)=0$ and $f_{y}(0,0)=0$.

*So, the explanation for $(1)$ is just by algebra because the limit in $(\star)$ is just
$$\lim_{(h,k)\to (0,0)}\frac{f(h,k)-\overbrace{f(0,0)-0(h-0)-0(k-0)}^{=0}}{\|(h,k)\|}$$

*The explanation for $(2)$ is again by algebra, because
$$\frac{f(h,k)}{\|(h,k)\|}=\frac{\frac{\sin^{2}(h+\sin(k))}{\sqrt[4]{h^{2}+k^{2}}}}{\|(h,k)\|}=\frac{\sin^{2}(h+\sin(k))}{\|(h,k)\|^{1/2}\cdot \|(h,k)\|}=\frac{\sin^{2}(h+\sin k)}{\|(h,k)\|^{3/2}},$$
since $\sqrt[4]{h^{2}+k^{2}}=\left(\left(h^{2}+k^{2}\right)^{1/2}\right)^{1/2}=\|(h,k)\|^{1/2}$.

*In $(3)$ we can use the fact $|\sin(x)|\leqslant |x|$ for all $x\in \mathbf{R}$ so $|\sin^{2}(h+\sin(k))|\leqslant(|h+\sin(k)|)^{2}\leqslant (|h|+|k|)^{2}$. Then, the last part should be \begin{align*}(|h|+|k|)^{2}&=|h|^{2}+2|h||k|+|k|^{2}\\&\leqslant |h|^{2}+|k|^{2}+|h|^{2}+|k|^{2},\quad (\star \star)\\&=2(|h|^{2}+|k|^{2})\\&=2\|(h,k)\|^{2}\\&<4\|(h,k)\|^{2}, \end{align*}
where the inequality $(\star\star)$ is follows of AM-GM with $\frac{|h|^{2}+|k|^{2}}{2}\geqslant \sqrt{|h|^{2}|k|^{2}}$. Hence,

\begin{align*}
\left|\frac{\sin^{2}(h+\sin(k))}{\|(h,k)\|^{3/2}}\right|&\leqslant \frac{(|h|+|k|)^{2}}{\|(h,k)\|^{3/2}},\\&\leqslant \frac{2\|(h,k)\|^{2}}{\|(h,k)\|^{3/2}},\\&=2\|(h,k)\|^{1/2},\quad \|(h,k)\|\not=0,\\&\longrightarrow 0,\quad (h,k)\longrightarrow 0
\end{align*}
Therefore since the partial derivatives there exists at $(0,0)$ and the limit given in $(\star)$ is $0$ and then $f$ is differentiable function at $(0,0)$.
